Chapter 12 Sequences; Induction; the Binomial Theorem
Chapter 12 Sequences; Induction; the Binomial Theorem
Chapter 12 Sequences; Induction; the Binomial Theorem
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<strong>Chapter</strong> <strong>12</strong><br />
<strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
Section <strong>12</strong>.1<br />
2−1 1<br />
2 2<br />
1. f ( 2)<br />
= = ; f ( 3)<br />
2. True<br />
3.<br />
4.<br />
nt ⋅<br />
3−1 2<br />
= =<br />
3 3<br />
⎛ r ⎞<br />
A= P⎜1+<br />
⎟<br />
⎝ n ⎠<br />
22 ⋅<br />
⎛ 0.04 ⎞<br />
= 1000⎜1+<br />
⎟<br />
⎝ 2 ⎠<br />
4<br />
= 1000( 1.02)<br />
= 1082.43<br />
After two years, <strong>the</strong> account will contain<br />
$1082.43.<br />
nt ⋅<br />
⎛ r ⎞<br />
A= P⎜1+<br />
⎟<br />
⎝ n ⎠<br />
⎛ 0.05 ⎞<br />
10,000 = P ⎜1+<br />
⎟<br />
⎝ <strong>12</strong> ⎠<br />
<strong>12</strong>⋅1<br />
⎛ 0.05 ⎞<br />
10,000 = P⎜1+<br />
⎟<br />
⎝ <strong>12</strong> ⎠<br />
10,000 = P( 1.051162)<br />
10,000<br />
= P<br />
1.051162<br />
9513.28 = P<br />
To have $10,000 at <strong>the</strong> end of one year, you need<br />
to invest $9513.28 now.<br />
5. sequence<br />
<strong>12</strong><br />
6. s 1 = 41 () − 1= 3; ( )<br />
4 4<br />
7. ∑( k)<br />
∑<br />
k= 1 k=<br />
1<br />
8. True<br />
s 4 = 4 4 − 1=<br />
15<br />
( + )<br />
4 4 1<br />
2 = 2 k = 2⋅ = 4( 5)<br />
= 20<br />
2<br />
9. True; a sequence is a function whose domain is<br />
<strong>the</strong> set of positive integers.<br />
11. 10! = 10987654321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 3,628,800<br />
<strong>12</strong>. 9! = 987654321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 362,880<br />
13.<br />
14.<br />
15.<br />
16.<br />
9! 9⋅8⋅7⋅6!<br />
= = 987 ⋅ ⋅ = 504<br />
6! 6!<br />
<strong>12</strong>! <strong>12</strong>⋅11⋅10!<br />
= = <strong>12</strong>⋅ 11 = 132<br />
10! 10!<br />
3!7! ⋅ 3217654!<br />
⋅ ⋅ ⋅ ⋅ ⋅ ⋅<br />
= = 321765 ⋅ ⋅ ⋅ ⋅ ⋅ = <strong>12</strong>60<br />
4! 4!<br />
5! ⋅8! 5⋅4⋅3! ⋅8!<br />
=<br />
3! 3!<br />
= 5487654321<br />
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅<br />
= 806,400<br />
17. s1 s2 s3 s4 s5<br />
18.<br />
= 1, = 2, = 3, = 4, = 5<br />
2 2 2<br />
1 2 3<br />
2 2<br />
4 = 4 + 1= 17, s5<br />
= 5 + 1=<br />
26<br />
s = 1 + 1= 2, s = 2 + 1= 5, s = 3 + 1=<br />
10,<br />
s<br />
1 1 2 2 1<br />
19. a1 = = , a2<br />
= = = ,<br />
1+ 2 3 2+<br />
2 4 2<br />
3 3 4 4 2<br />
a3 = = , a4<br />
= = = ,<br />
3+ 2 5 4+<br />
2 6 3<br />
5 5<br />
a 5 = =<br />
5+<br />
2 7<br />
21 ⋅ + 1 3 22 ⋅ + 1 5<br />
20. b1 = = , b2<br />
= = ,<br />
21 ⋅ 2 22 ⋅ 4<br />
23 ⋅ + 1 7 24 ⋅ + 1 9<br />
b3 = = , b4<br />
= = ,<br />
23 ⋅ 6 24 ⋅ 8<br />
25 ⋅ + 1 11<br />
b5<br />
= =<br />
25 ⋅ 10<br />
21.<br />
11 + 2 21 + 2<br />
1 c2<br />
3+ 1 2 4+<br />
1 2<br />
3 c4<br />
5+<br />
1 2<br />
c = ( − 1) (1 ) = 1, = ( − 1) (2 ) =−4,<br />
c = ( − 1) (3 ) = 9, = ( − 1) (4 ) = −16,<br />
c 5 = ( − 1) (5 ) = 25<br />
10. True;<br />
2<br />
∑<br />
k = 1<br />
( + )<br />
2 2 1<br />
k = = 3<br />
2<br />
<strong>12</strong>35<br />
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as <strong>the</strong>y currently<br />
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
22.<br />
23.<br />
24.<br />
25.<br />
26.<br />
11 − ⎛ 1 ⎞<br />
d1<br />
= ( − 1) ⎜ ⎟=<br />
1,<br />
⎝21 ⋅ −1⎠<br />
2−1⎛<br />
2 ⎞ 2<br />
d2<br />
= ( − 1) ⎜ ⎟= − ,<br />
⎝2⋅2−1⎠<br />
3<br />
3−1⎛<br />
3 ⎞ 3<br />
d3<br />
= ( − 1) ⎜ ⎟=<br />
,<br />
⎝23 ⋅ −1⎠<br />
5<br />
d<br />
d<br />
4<br />
5<br />
4−1⎛<br />
4 ⎞ 4<br />
= ( − 1) ⎜ ⎟= − ,<br />
⎝2⋅4−1⎠<br />
7<br />
5−1⎛<br />
5 ⎞ 5<br />
= ( − 1) ⎜ ⎟=<br />
⎝25 ⋅ −1⎠<br />
9<br />
1 2<br />
2 2 1 2 4 2<br />
s1 = = = , s<br />
1 2 = = = ,<br />
2<br />
3 + 1 4 2 3 + 1 10 5<br />
s<br />
3 4<br />
2 8 2 2 16 8<br />
= = = , s = = = ,<br />
3 + 1 28 7 3 + 1 82 41<br />
3 3 4 4<br />
s<br />
5 5<br />
5<br />
2 32 8<br />
= = =<br />
3 + 1 244 61<br />
1 2<br />
⎛4⎞ 4 ⎛4⎞<br />
16<br />
s1 = ⎜ ⎟ = , s2<br />
= ⎜ ⎟ = ,<br />
⎝3⎠ 3 ⎝3⎠<br />
9<br />
s<br />
s<br />
3 4<br />
3 4<br />
5<br />
⎛4 ⎞ 64 ⎛4 ⎞ 256<br />
= ⎜ ⎟ = , s = ⎜ ⎟ = ,<br />
⎝3⎠ 27 ⎝3⎠<br />
81<br />
5<br />
⎛4⎞<br />
1024<br />
= ⎜ ⎟ =<br />
⎝3⎠<br />
243<br />
1<br />
( −1) −1 1<br />
t1<br />
= = = − ,<br />
(1+ 1)(1 + 2) 2 ⋅3 6<br />
t<br />
t<br />
t<br />
t<br />
2<br />
3<br />
4<br />
5<br />
2<br />
( −1) 1 1<br />
= = = ,<br />
(2 + 1)(2 + 2) 3⋅4 <strong>12</strong><br />
3<br />
( −1) −1 1<br />
= = = − ,<br />
(3 + 1)(3 + 2) 4⋅5 20<br />
4<br />
( −1) 1 1<br />
= = = ,<br />
(4 + 1)(4 + 2) 5⋅6 30<br />
5<br />
( −1) −1 1<br />
= = =−<br />
(5 + 1)(5 + 2) 6⋅7 42<br />
1 2 3<br />
3 3 3 9 3 27<br />
a1 = = = 3, a2 = = , a3<br />
= = = 9,<br />
1 1 2 2 3 3<br />
a<br />
4 5<br />
3 81 3 243<br />
= = , a = =<br />
4 4 5 5<br />
4 5<br />
1 1 2 3 4 5<br />
e e e e e e<br />
27. b1 = = , b<br />
1 2 = , b<br />
2 3 = , b<br />
3 4 = , b<br />
4 5 = 5<br />
28.<br />
2 2 2<br />
1 1 2 2 3 3<br />
1 1 2 3 9<br />
c = = , c = = 1, c = = ,<br />
2 2 2 2 8<br />
2 2<br />
4 16 5 25<br />
c4 = = = 1, c<br />
4 5 = =<br />
5<br />
2 16 2 32<br />
29. Answers may vary. One possibility follows:<br />
Each term is a fraction with <strong>the</strong> numerator equal<br />
to <strong>the</strong> term number and <strong>the</strong> denominator equal to<br />
one more than <strong>the</strong> term number.<br />
n<br />
an<br />
=<br />
n + 1<br />
30. Answers may vary. One possibility follows:<br />
Each term is a fraction with <strong>the</strong> numerator equal<br />
to 1 and <strong>the</strong> denominator equal to <strong>the</strong> product of<br />
<strong>the</strong> term number and one more than <strong>the</strong> term<br />
number.<br />
1<br />
a n =<br />
n n+<br />
1<br />
( )<br />
31. Answers may vary. One possibility follows:<br />
Each term is a fraction with <strong>the</strong> numerator equal<br />
to 1 and <strong>the</strong> denominator equal to a power of 2.<br />
The power is equal to one less than <strong>the</strong> term<br />
number.<br />
1<br />
a n =<br />
n−1<br />
2<br />
32. Answers may vary. One possibility follows:<br />
Each term is equal to a fraction with <strong>the</strong><br />
numerator equal to a power of 2 and <strong>the</strong><br />
denominator equal to a power of 3. Both powers<br />
are equal to <strong>the</strong> term number. Since <strong>the</strong> powers<br />
are <strong>the</strong> same, we can use rules for exponents to<br />
write each term as a power of 2 3 .<br />
a n<br />
⎛2<br />
⎞<br />
= ⎜ ⎟<br />
⎝3<br />
⎠<br />
n<br />
33. Answers may vary. One possibility follows:<br />
The terms form an alternating sequence. Ignoring<br />
<strong>the</strong> sign, each term always contains a 1. The sign<br />
alternates by raising − 1 to a power. Since <strong>the</strong><br />
first term is positive, we use n − 1 as <strong>the</strong> power.<br />
1<br />
1 n −<br />
= −<br />
a n<br />
( )<br />
<strong>12</strong>36<br />
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as <strong>the</strong>y currently<br />
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
Section <strong>12</strong>.1: <strong>Sequences</strong><br />
34. Answers may vary. One possibility follows:<br />
The terms appear to alternate between whole<br />
numbers and fractions. If we write <strong>the</strong> whole<br />
numbers as fractions (e.g. 1 = 1 , 3 = 3 , etc.), we<br />
1 1<br />
see that each term consists of a 1 and <strong>the</strong> term<br />
number. When n is odd, <strong>the</strong> numerator is n and <strong>the</strong><br />
denominator is 1. When n is even, <strong>the</strong> numerator is<br />
1 and <strong>the</strong> denominator is n. This alternating<br />
behavior occurs if we have a power that alternates<br />
sign. The alternating sign is obtained by using<br />
( − )<br />
1 n+1<br />
. Thus, we get<br />
an<br />
( −<br />
= n<br />
)<br />
1<br />
1 n +<br />
35. Answers may vary. One possibility follows:<br />
The terms (ignoring <strong>the</strong> sign) are equal to <strong>the</strong><br />
term number. The alternating sign is obtained by<br />
using ( − 1)<br />
n+1<br />
.<br />
an<br />
( )<br />
1<br />
1 n +<br />
= − ⋅ n<br />
36. Answers may vary. One possibility follows:<br />
Here again we have alternating signs which will<br />
be taken care of by using ( − 1)<br />
n+1<br />
. The rest of <strong>the</strong><br />
term is twice <strong>the</strong> term number.<br />
an<br />
( )<br />
n+<br />
1<br />
= −1 ⋅ 2n<br />
37. a1 a2 a3<br />
= 2, = 3 + 2 = 5, = 3 + 5 = 8,<br />
a = 3 + 8 = 11, a = 3+ 11 = 14<br />
4 5<br />
a = 3, a = 4 − 3 = 1, a = 4 − 1 = 3,<br />
a = 4− 3= 1, a = 4− 1=<br />
3<br />
38. 1 2 3<br />
4 5<br />
39. a1 a2 a3<br />
=− 2, = 2 + ( − 2) = 0, = 3 + 0 = 3,<br />
a = 4+ 3= 7, a = 5+ 7 = <strong>12</strong><br />
4 5<br />
40. a1 a2 a3<br />
= 1, = 2 − 1 = 1, = 3 − 1 = 2,<br />
a = 4− 2= 2, a = 5− 2=<br />
3<br />
4 5<br />
a = 5, a = 2⋅ 5 = 10, a = 2⋅ 10 = 20,<br />
a = 220 ⋅ = 40, a = 240 ⋅ = 80<br />
41. 1 2 3<br />
4 5<br />
42. a1 a2 a3<br />
= 2, = − 2, = −( − 2) = 2,<br />
a =− 2, a =−− ( 2) = 2<br />
4 5<br />
3 1 1<br />
3<br />
43. 2 1 2 1 8 1<br />
a1= 3, a2 = , a3 = = , a4<br />
= = , a5<br />
= =<br />
2 3 2 4 8 5 40<br />
44. a1 = − 2, a2<br />
= 2 + 3( − 2) =−4,<br />
a3 = 3 + 3( − 4) = − 9, a4<br />
= 4 + 3( − 9) =−23,<br />
a 5 = 5+ 3( − 23) =− 64<br />
= 1, = 2, = 2⋅ 1 = 2, = 2⋅ 2 = 4,<br />
a = 42 ⋅ = 8<br />
45. a1 a2 a3 a4<br />
5<br />
46. a1 a2 a3<br />
= − 1, = 1, =− 1+ 3⋅ 1 = 2,<br />
a = 1+ 4⋅ 2= 9, a = 2+ 5⋅ 9=<br />
47<br />
4 5<br />
a = A, a = A+ d, a = ( A+ d) + d = A+<br />
2 d,<br />
a4<br />
= ( A+ 2 d) + d = A+<br />
3 d,<br />
a = ( A+ 3 d) + d = A+<br />
4d<br />
47. 1 2 3<br />
48.<br />
5<br />
2<br />
1 = , 2 = , 3 = ( ) = ,<br />
2 3 3 4<br />
4 = = 5 = =<br />
a A a rA a r rA r A<br />
( ) , ( )<br />
a r r A r A a r r A r A<br />
49. a1 a2 a3<br />
a<br />
4<br />
= 2, = 2+ 2, = 2+ 2+<br />
2,<br />
= 2+ 2+ 2+<br />
2 ,<br />
a 5 = 2+ 2+ 2+ 2+<br />
2<br />
2<br />
2<br />
= 2, = , =<br />
2<br />
,<br />
2 2<br />
50. a1 a2 a3<br />
n<br />
a<br />
2<br />
2 2<br />
2 2<br />
=<br />
2<br />
, a =<br />
2<br />
2 2<br />
4 5<br />
51. ∑ ( k+ 2) = 3+ 4+ 5+ 6+ 7+⋅⋅⋅+ ( n+<br />
2)<br />
k = 1<br />
n<br />
52. ∑ (2k<br />
+ 1) = 3+ 5 + 7 + 9 +⋅⋅⋅+ ( 2n+<br />
1)<br />
53.<br />
k = 1<br />
n<br />
∑<br />
k = 1<br />
n<br />
2 2<br />
k 1 9 25 49 n<br />
= + 2 + + 8 + + 18 + + 32 +⋅⋅⋅+<br />
2 2 2 2 2 2<br />
2 2<br />
54. ∑ ( k+ 1) = 4+ 9+ 16+ 25+ 36+⋅⋅⋅+ ( n+<br />
1)<br />
k = 1<br />
<strong>12</strong>37<br />
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as <strong>the</strong>y currently<br />
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
55.<br />
56.<br />
57.<br />
n<br />
1 1 1 1 1<br />
∑ = 1+ + + + +<br />
3<br />
k 3 9 27 3<br />
n<br />
k = 0<br />
n<br />
⎛3⎞ 3 9 ⎛3⎞<br />
∑ ⎜ ⎟ = 1+ + + + ⎜ ⎟<br />
⎝2⎠ 2 4 ⎝2⎠<br />
k = 0<br />
n−1<br />
k = 0<br />
k<br />
1 1 1 1 1<br />
∑ = + + + +<br />
k+<br />
1<br />
3 3 9 27<br />
n<br />
3<br />
n−1<br />
58. ∑ (2k<br />
+ 1) = 1+ 3+ 5+ 7 + + ( 2( n− 1) + 1)<br />
59.<br />
60.<br />
k = 0<br />
n<br />
k = 2<br />
= 1+ 3+ 5+ 7 + + (2n<br />
−1)<br />
k<br />
n<br />
∑ ( − 1) lnk<br />
= ln2− ln3+ ln4 − + ( −1) lnn<br />
n<br />
∑<br />
k = 3<br />
k+<br />
1 k<br />
( −1) 2<br />
4 3 5 4 6 5 n+<br />
1 n<br />
= ( − 1) 2 + ( − 1) 2 + ( − 1) 2 + + ( −1) 2<br />
3 4 5 6 n+<br />
1 n<br />
= 2 − 2 + 2 − 2 + + ( −1) 2<br />
n+<br />
1 n<br />
= 8− 16+ 32− 64 + ... + ( −1) 2<br />
61. Answers may vary. One possibility follows:<br />
20<br />
1+ 2+ 3+ + 20= ∑ k<br />
k = 1<br />
62. Answers may vary. One possibility follows:<br />
8<br />
3 3 3 3 3<br />
1 + 2 + 3 + + 8 = ∑ k<br />
k = 1<br />
63. Answers may vary. One possibility follows:<br />
1 2 3 13<br />
13<br />
k<br />
+ + + + = ∑<br />
2 3 4 13+ 1 k + 1<br />
k = 1<br />
64. Answers may vary. One possibility follows:<br />
<strong>12</strong><br />
1+ 3 + 5 + 7 + + 2(<strong>12</strong>) − 1 = (2k<br />
−1)<br />
[ ]<br />
n<br />
∑<br />
k = 1<br />
65. Answers may vary. One possibility follows:<br />
1 1 1<br />
6<br />
6 ⎛ 1 ⎞ k ⎛ 1 ⎞<br />
1 − + − + + ( − 1) ⎜ ( 1)<br />
3 9 27<br />
6 ⎟= ∑ − ⎜ k ⎟<br />
⎝3 ⎠ ⎝3<br />
⎠<br />
k = 0<br />
66. Answers may vary. One possibility follows:<br />
11<br />
2 4 8<br />
11<br />
11+ 1 ⎛2⎞ k + 1 ⎛2⎞<br />
− + + + ( − 1) ⎜ ⎟ = ∑ ( −1)<br />
⎜ ⎟<br />
3 9 27<br />
⎝3⎠ ⎝3⎠<br />
k = 1<br />
k<br />
67. Answers may vary. One possibility follows:<br />
2 3<br />
n k<br />
3 3 3 3<br />
3 + n<br />
2 + 3<br />
+ + n<br />
= ∑ k<br />
k = 1<br />
68. Answers may vary. One possibility follows:<br />
1 2 3 n<br />
n<br />
k<br />
+ + + + =<br />
e<br />
2 3<br />
n<br />
∑<br />
k<br />
e e e e<br />
k = 1<br />
69. Answers may vary. One possibility follows:<br />
n<br />
a+ ( a+ d) + ( a+ 2 d) + + ( a+ nd) = ( a+<br />
kd)<br />
n<br />
∑<br />
k = 1<br />
( )<br />
or = ( a+ k −1 d)<br />
∑<br />
k = 0<br />
70. Answers may vary. One possibility follows:<br />
n<br />
2 n−1 k−1<br />
a+ ar+ ar + + ar = ∑ ar<br />
or<br />
n−1<br />
∑<br />
k = 0<br />
ar<br />
k<br />
k = 1<br />
40<br />
71. ( )<br />
72.<br />
73.<br />
74.<br />
75.<br />
∑ 5= 5 <br />
+ 5+ 5+⋅⋅⋅+ 5= 40 5 = 200<br />
k = 1 40 times<br />
50<br />
∑ 8 = 8 <br />
+ 8 + 8 +⋅⋅⋅+ 8 = 50(8) = 400<br />
k = 1 50 times<br />
40<br />
∑<br />
k = 1<br />
( + )<br />
40 40 1<br />
k = = 20( 41)<br />
= 820<br />
2<br />
24 24<br />
k= 1 k=<br />
1<br />
( + )<br />
24 24 1<br />
( − k) = − k = − = −300<br />
2<br />
∑ ∑<br />
20 20 20 20 20<br />
∑ ∑ ∑ ∑ ∑<br />
(5k+ 3) = (5 k) + 3 = 5 k+<br />
3<br />
k= 1 k= 1 k= 1 k= 1 k=<br />
1<br />
( + )<br />
⎛20 20 1 ⎞<br />
= 5⎜<br />
⎟+<br />
3 20<br />
⎝ 2 ⎠<br />
= 1050 + 60 = 1110<br />
( )<br />
26 26 26 26 26<br />
76. ∑ ( k − ) = ∑( k)<br />
− ∑ = ∑k−∑<br />
3 7 3 7 3 7<br />
k= 1 k= 1 k= 1 k= 1 k=<br />
1<br />
( + )<br />
⎛26 26 1 ⎞<br />
= 3⎜<br />
⎟−7 26<br />
⎝ 2 ⎠<br />
= 1053 − 182 = 871<br />
( )<br />
<strong>12</strong>38<br />
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Section <strong>12</strong>.1: <strong>Sequences</strong><br />
16 16 16<br />
2 2<br />
∑ ∑ ∑<br />
77. ( k )<br />
+ 4 = k + 4<br />
k= 1 k= 1 k=<br />
1<br />
( + )( ⋅ + )<br />
16 16 1 2 16 1<br />
= + 416<br />
6<br />
= 1496 + 64 = 1560<br />
( )<br />
14 14<br />
78. ∑ ( k<br />
2 − 4) = ( 0 2 − 4) + ∑( k<br />
2 −4)<br />
79.<br />
k= 0 k=<br />
1<br />
14 14<br />
2<br />
∑<br />
= k −<br />
∑<br />
k= 1 k=<br />
1<br />
4<br />
( + )( ⋅ + )<br />
14 14 1 2 14 1<br />
=− 4+ −4 14<br />
6<br />
=− 4 + 1015− 64 = 955<br />
⎡ ⎤<br />
2k = 2 2k = 2⎢<br />
k − k⎥<br />
⎣ ⎦<br />
60 60 60 9<br />
∑ ∑ ∑ ∑<br />
k= 10 k= 10 k= 1 k=<br />
1<br />
( + ) ( + )<br />
⎡60 60 1 9 9 1 ⎤<br />
= 2 ⎢ − ⎥<br />
⎣ 2 2 ⎦<br />
= 2 1830 − 45 = 3570<br />
[ ]<br />
⎡ ⎤<br />
k k ⎢ k k⎥<br />
⎣ ⎦<br />
40 40 40 7<br />
80. ∑( − 3 ) =− 3∑ =−3<br />
∑ −∑<br />
k= 8 k= 8 k= 1 k=<br />
1<br />
( + ) ( + )<br />
⎡40 40 1 7 7 1 ⎤<br />
=−3⎢<br />
− ⎥<br />
⎣ 2 2 ⎦<br />
=−3 820 − 28 =−2376<br />
[ ]<br />
( )<br />
From <strong>the</strong> table we see that <strong>the</strong> balance is<br />
below $2000 after 14 payments have been<br />
made. The balance <strong>the</strong>n is $1953.70.<br />
c. Scrolling down <strong>the</strong> table, we find that<br />
balance is paid off in <strong>the</strong> 36th month. The<br />
last payment is $83.78. There are 35<br />
payments of $100 and <strong>the</strong> last payment of<br />
$83.78. The total amount paid is: 35(100) +<br />
83.78(1.01) = $3584.62. (we have to add<br />
<strong>the</strong> interest for <strong>the</strong> last month).<br />
d. The interest expense is:<br />
3584.62 – 3000.00 = $584.62<br />
81.<br />
82.<br />
20 20 4<br />
3 3 3<br />
∑ ∑ ∑<br />
k = k − k<br />
k= 5 k= 1 k=<br />
1<br />
( + ) ( + )<br />
2 2<br />
2 2<br />
⎡20 20 1 ⎤ ⎡4 4 1 ⎤<br />
= ⎢ ⎥ −⎢ ⎥<br />
⎣ 2 ⎦ ⎣ 2 ⎦<br />
= 210 − 10 = 44,000<br />
24 24 3<br />
3 3 3<br />
∑ ∑ ∑<br />
k = k − k<br />
k= 4 k= 1 k=<br />
1<br />
( + ) ( + )<br />
2 2<br />
2 2<br />
⎡24 24 1 ⎤ ⎡3 3 1 ⎤<br />
= ⎢ ⎥ −⎢ ⎥<br />
⎣ 2 ⎦ ⎣ 2 ⎦<br />
= 300 − 6 = 89,964<br />
84. a. B 1 = 1.005(18500) − 534.47 = $18,058.03<br />
b. Put <strong>the</strong> graphing utility in SEQuence mode.<br />
Enter Y= as follows, <strong>the</strong>n examine <strong>the</strong><br />
TABLE:<br />
83. a. B 1 = 1.01(3000) − 100 = $2930<br />
b. Put <strong>the</strong> graphing utility in SEQuence mode.<br />
Enter Y= as follows, <strong>the</strong>n examine <strong>the</strong><br />
TABLE:<br />
From <strong>the</strong> table we see that <strong>the</strong> balance is<br />
below $10,000 after 19 payments have been<br />
made. The balance <strong>the</strong>n is $9713.76.<br />
<strong>12</strong>39<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
c. Scrolling down <strong>the</strong> table, we find that<br />
balance is paid off in <strong>the</strong> 39th month. The<br />
last payment is $54.18. There are 38<br />
payments of $534.47 and <strong>the</strong> last payment<br />
of $54.18 plus interest. The total amount<br />
paid is: 38(534.47) + 54.18(1.005) =<br />
$20,364.31.<br />
d. The interest expense is:<br />
20,364.31 – 18,500.00 = $1864.31<br />
85. a. p1<br />
= 1.03(2000) + 20 = 2080;<br />
p2<br />
= 1.03(2080) + 20 = 2162.4<br />
There are approximately 2162 trout in <strong>the</strong><br />
pond at <strong>the</strong> end of <strong>the</strong> second month.<br />
b. Scrolling down <strong>the</strong> table, we find <strong>the</strong> trout<br />
population exceeds 5000 at <strong>the</strong> end of <strong>the</strong><br />
26th month when <strong>the</strong> population is 5084.<br />
c. The equilibrium level of pollution occurs<br />
when x = 0.9x+ 15 . That is, when x = 150<br />
tons.<br />
x = 0.9x+<br />
15<br />
0.1x<br />
= 15<br />
x = 150<br />
87. a. Since <strong>the</strong> fund returns 8% compound<br />
annually, this is equivalent to a return of 2%<br />
each quarter. Defining a recursive<br />
sequence, we have:<br />
a0 = 0, an<br />
= 1.02an<br />
− 1+<br />
500<br />
b. Insert <strong>the</strong> formulas in your graphing utility<br />
and use <strong>the</strong> table feature to find when <strong>the</strong><br />
value of <strong>the</strong> account will exceed $100,000:<br />
86. a. p1<br />
= 0.9(250) + 15 = 240;<br />
p2<br />
= 0.9(240) + 15 = 231<br />
There are 240 tons of pollutants at <strong>the</strong> end of<br />
<strong>the</strong> first year, and 231 tons of pollutants at<br />
<strong>the</strong> end of <strong>the</strong> second year.<br />
b. Scrolling down <strong>the</strong> table, we display <strong>the</strong><br />
pollutant levels for <strong>the</strong> next 20 years.<br />
In <strong>the</strong> 82nd quarter (approximately May<br />
2019) <strong>the</strong> value of <strong>the</strong> account will exceed<br />
$100,000 with a value of $101,810.<br />
<strong>12</strong>40<br />
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Section <strong>12</strong>.1: <strong>Sequences</strong><br />
c. Find <strong>the</strong> value of <strong>the</strong> account in 25 years or<br />
100 quarters:<br />
c. Enter <strong>the</strong> recursive formula in Y= and create<br />
<strong>the</strong> table:<br />
The value of <strong>the</strong> account will be $156,116.15.<br />
88. a. Since <strong>the</strong> fund returns 6% compounded<br />
annually, this is equivalent to a return of<br />
0.5% each month. Defining a recursive<br />
sequence, we have:<br />
a0 = 0, an<br />
= 1.005an<br />
− 1+<br />
45<br />
b. Insert <strong>the</strong> formulas in your graphing utility<br />
and use <strong>the</strong> table feature to find when <strong>the</strong><br />
value of <strong>the</strong> account will exceed $4,000:<br />
d. Scroll through <strong>the</strong> table:<br />
After 58 payments have been made, <strong>the</strong><br />
balance is below $140,000. The balance is<br />
about $139,981.<br />
e. Scroll through <strong>the</strong> table:<br />
In <strong>the</strong> 74th month (February 2006) <strong>the</strong> value<br />
of <strong>the</strong> account will exceed $4,000 with a<br />
value of about $4,017.60.<br />
c. Find <strong>the</strong> value of <strong>the</strong> account in 16 years or<br />
192 months:<br />
The value of <strong>the</strong> account will be<br />
approximately $14,449.11.<br />
89. a. Since <strong>the</strong> interest rate is 6% per annum<br />
compounded monthly, this is equivalent to a<br />
rate of 0.5% each month. Defining a<br />
recursive sequence, we have:<br />
a0 = 150,000, an<br />
= 1.005an<br />
− 1−<br />
899.33<br />
b. 1.005(150,000) − 899.33 = $149,850.67<br />
The loan will be paid off at <strong>the</strong> end of 360<br />
months or 30 years.<br />
Total amount paid = (359)($899.33) +<br />
$890.65(1.005) = $323,754.57.<br />
f. The total interest expense is <strong>the</strong> difference<br />
of <strong>the</strong> total of <strong>the</strong> payments and <strong>the</strong> original<br />
loan: 323,754.57 − 150,000 = $173,754.57<br />
g. (a) Since <strong>the</strong> interest rate is 6% per annum<br />
compounded monthly, this is equivalent to<br />
a rate of 0.5% each month. Defining a<br />
recursive sequence, we have:<br />
a0 = 150,000, an<br />
= 1.005an<br />
− 1−<br />
999.33<br />
(b) 1.005(150,000) − 999.33 = $149,750.67<br />
<strong>12</strong>41<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
(c) Enter <strong>the</strong> recursive formula in Y= and<br />
create <strong>the</strong> table:<br />
b.<br />
⎛ 0.065 ⎞<br />
⎜1 + ⎟(<strong>12</strong>0,000) −758.48<br />
= $119,891.52<br />
⎝ <strong>12</strong> ⎠<br />
c. Enter <strong>the</strong> recursive formula in Y= and create<br />
<strong>the</strong> table:<br />
(d) Scroll through <strong>the</strong> table:<br />
d. Scroll through <strong>the</strong> table:<br />
After 37 payments have been made, <strong>the</strong><br />
balance is below $140,000. The<br />
balance is $139,894.<br />
(e) Scroll through <strong>the</strong> table:<br />
After <strong>12</strong>9 payments have been made, <strong>the</strong><br />
balance is below $100,000. The balance is<br />
about $99,824.<br />
e. Scroll through <strong>the</strong> table:<br />
The loan will be paid off at <strong>the</strong> end of<br />
279 months or 23 years and 3 months.<br />
Total amount paid = (278)($999.33) +<br />
353.69(1.005) = $278,169.20<br />
(f) The total interest expense is <strong>the</strong><br />
difference of <strong>the</strong> total of <strong>the</strong> payments<br />
and <strong>the</strong> original loan:<br />
278,169.20 − 150,000 = $<strong>12</strong>8,169.20<br />
h. Yes, if <strong>the</strong>y can afford <strong>the</strong> additional<br />
monthly payment. They would save<br />
$44,586.07 in interest payments by paying<br />
<strong>the</strong> loan off sooner.<br />
90. a. Since <strong>the</strong> interest rate is 6.5% per annum<br />
compounded at a rate of (6.5/<strong>12</strong>)% each<br />
month. Defining a recursive sequence, we<br />
have:<br />
a0<br />
= <strong>12</strong>0,000,<br />
⎛ 0.065 ⎞<br />
an<br />
= ⎜1+ ⎟an<br />
− 1 −758.48<br />
⎝ <strong>12</strong> ⎠<br />
The loan will be paid in <strong>the</strong> 360th month i.e.<br />
after 30 years.<br />
Total amount paid = (359)(758.48) +<br />
756.19(1+0.065/<strong>12</strong>) = $273,054.60.<br />
f. The total interest expense is <strong>the</strong> difference<br />
of <strong>the</strong> total of <strong>the</strong> payments and <strong>the</strong> original<br />
loan: 273,054.60 − <strong>12</strong>0,000 = $153,054.60<br />
g. (a) Since <strong>the</strong> interest rate is 6.5% per<br />
annum compounded monthly, this is<br />
equivalent to a rate of (6.5/<strong>12</strong>)% each<br />
month. Defining a recursive sequence,<br />
we have:<br />
a0<br />
= <strong>12</strong>0,000,<br />
⎛ 0.065 ⎞<br />
an<br />
= ⎜1+ ⎟an<br />
− 1 −858.48<br />
⎝ <strong>12</strong> ⎠<br />
<strong>12</strong>42<br />
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Section <strong>12</strong>.1: <strong>Sequences</strong><br />
(b)<br />
⎛ 0.065 ⎞<br />
⎜1 + ⎟(<strong>12</strong>0, 000) −858.48<br />
⎝ <strong>12</strong> ⎠<br />
= $119,791.52<br />
(c) Enter <strong>the</strong> recursive formula in Y= and<br />
create <strong>the</strong> table:<br />
(d) Scroll through <strong>the</strong> table:<br />
After 78 payments have been made, <strong>the</strong><br />
balance is below $100,000. The<br />
balance is $99,831.<br />
(e) Scroll through <strong>the</strong> table:<br />
The loan will be paid off at <strong>the</strong> end of<br />
262 months or 21 years and 10 months.<br />
Total amount paid = (261)(858.48) +<br />
(851.23)(1+0.065/<strong>12</strong>) = $224,919.<strong>12</strong>.<br />
(f) The total interest expense is <strong>the</strong><br />
difference of <strong>the</strong> total of <strong>the</strong> payments<br />
and <strong>the</strong> original loan:<br />
224,919.<strong>12</strong> − <strong>12</strong>0,000 = $104,919.<strong>12</strong><br />
h. Yes, if <strong>the</strong>y can afford <strong>the</strong> additional<br />
monthly payment. They would save<br />
$48,238.49 in interest payments by paying<br />
<strong>the</strong> loan off sooner.<br />
91. a1 a2 a3 a4 a5<br />
92. a.<br />
= 1, = 1, = 2, = 3, = 5,<br />
a6 = 8, a7 = 13, a8 = 21, an = an−1+<br />
an−2<br />
a8 = a7 + a6 = 13+ 8 = 21<br />
After 7 months <strong>the</strong>re are 21 mature pairs of<br />
rabbits.<br />
b.<br />
( 1+ 5) −( 1−<br />
5)<br />
1 1<br />
u1 =<br />
1<br />
2 5<br />
1+ 5− 1+<br />
5 2 5<br />
= = = 1<br />
2 5 2 5<br />
u<br />
u<br />
=<br />
( 1+ 5) −( 1−<br />
5)<br />
2 2<br />
2 2<br />
=<br />
2 5<br />
1+ 2 5+ 5− 1+ 2 5−5 4 5<br />
= = = 1<br />
4 5 4 5<br />
n+<br />
1<br />
+ u<br />
n<br />
n n n n<br />
+ 1 + 1<br />
( 1+ 5) −( 1− 5) ( 1+ 5) −( 1−<br />
5)<br />
+<br />
n+<br />
1<br />
n<br />
2 5 2 5<br />
+ 1 + 1<br />
( 1+ 5) −( 1− 5) + 2( 1+ 5) −2( 1−<br />
5)<br />
n n n n<br />
=<br />
n+<br />
1<br />
2 5<br />
n<br />
n<br />
( 1+ 5) ⎡1+ 5 + 2⎤−( 1− 5)<br />
⎡1− 5+<br />
2⎤<br />
=<br />
⎣ ⎦ ⎣ ⎦<br />
n+<br />
1<br />
2 5<br />
n<br />
n<br />
( 1+ 5) ⎡3+ 5⎤−( 1− 5)<br />
⎡3−<br />
5⎤<br />
=<br />
⎣ ⎦ ⎣ ⎦<br />
n+<br />
1<br />
2 5<br />
n+ 2 ( 3+ 5)<br />
n+<br />
2 ( 3−<br />
5)<br />
( 1+ 5)<br />
−<br />
2 ( 1−<br />
5)<br />
2<br />
( 1+ 5)<br />
( 1−<br />
5)<br />
=<br />
n+<br />
1<br />
2 5<br />
n+ 2 ( 3+ 5)<br />
n+<br />
2 ( 3−<br />
5)<br />
( 1+ 5)<br />
−( 1−<br />
5)<br />
( 6+ 2 5)<br />
( 6−2 5)<br />
=<br />
n+<br />
1<br />
2 5<br />
n+ 2 1 n+<br />
2 1<br />
( 1+ 5) −( 1−<br />
5)<br />
=<br />
2 2<br />
n+<br />
1<br />
2 5<br />
=<br />
n<br />
( 1+ 5) −( 1−<br />
5)<br />
= u n + 2<br />
+ 2 n+<br />
2<br />
2<br />
n+<br />
2<br />
5<br />
c. Since u = 1, u = 1, u = u + u , { u }<br />
1 2 n+ 2 n+<br />
1 n n<br />
is <strong>the</strong> Fibonacci sequence.<br />
<strong>12</strong>43<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
93. 1, 1, 2, 3, 5, 8, 13<br />
This is <strong>the</strong> Fibonacci sequence.<br />
94. a. u1 = 1, u2 = 1, u3 = 2, u4 = 3, u5<br />
= 5,<br />
u6 = 8, u7 = 13, u8 = 21, u9 = 34, u10<br />
= 55,<br />
u = 89<br />
b.<br />
11<br />
u2 1 u3<br />
2 u4<br />
3<br />
= = 1, = = 2, = = 1.5,<br />
u1 1 u2 1 u3<br />
2<br />
u5 5 u6<br />
8<br />
= ≈ 1.67, = = 1.6,<br />
u4 3 u5<br />
5<br />
u7 13 u8<br />
21<br />
= = 1.625, = ≈1.615,<br />
u6 8 u7<br />
13<br />
u9 34 u10<br />
55<br />
= ≈ 1.619, = ≈1.618,<br />
u8 21 u9<br />
34<br />
u11<br />
89<br />
= ≈1.618<br />
u 55<br />
10<br />
⎛<br />
1+<br />
5⎞<br />
c. 1.618 ⎜The exact value is ⎟<br />
⎝<br />
2 ⎠<br />
u1 1 u2<br />
1 u3<br />
2<br />
d. = = 1, = = 0.5, = ≈ 0.667,<br />
u2 1 u3 2 u4<br />
3<br />
u4<br />
3 u5<br />
5<br />
= = 0.6, = = 0.625,<br />
u5 5 u6<br />
8<br />
u6 8 u7<br />
13<br />
= ≈ 0.615, = ≈0.619,<br />
u7 13 u8<br />
21<br />
u8 21 u9<br />
34<br />
= ≈ 0.618, = ≈0.618,<br />
u9 34 u10<br />
55<br />
u10<br />
55<br />
= ≈0.618<br />
u 89<br />
11<br />
1.3<br />
c. f ( ) e<br />
1.3 = ≈ 3.669296668<br />
d. It will take n = <strong>12</strong> to approximate<br />
f<br />
1.3<br />
( 1.3) e<br />
96. a. ( 2.4)<br />
= correct to 8 decimal places.<br />
3<br />
( 2.4)<br />
∑<br />
2.4<br />
f e − −<br />
− = ≈<br />
b. ( 2.4)<br />
k = 0<br />
k!<br />
( −2.4) ( −2.4) ( −2.4) ( −2.4)<br />
+ + +<br />
0 1 2 3<br />
=<br />
0! 1! 2! 3!<br />
=−0.824<br />
2.4 ( 2.4)<br />
f e − −<br />
− = ≈<br />
2.4<br />
c. f ( ) e −<br />
6<br />
∑<br />
k = 0<br />
k!<br />
( −2.4) ( −2.4) ( −2.4)<br />
= + + ... +<br />
0 1 6<br />
0! 1! 6!<br />
= 0.1602688<br />
− 2.4 = ≈ 0.0907179533<br />
k<br />
k<br />
e. 0.618<br />
⎛<br />
2 ⎞<br />
⎜The exact value is ⎟<br />
⎝<br />
1 + 5 ⎠<br />
d. It will take n = 17 to approximate<br />
f<br />
2.4<br />
( 2.4) e −<br />
− = correct to 8 decimal places.<br />
95. a. f ( 1.3)<br />
b. f ( 1.3)<br />
4<br />
k<br />
1.3 1.3<br />
∑ k!<br />
k = 0<br />
= e ≈<br />
0 1 4<br />
1.3 1.3 1.3<br />
= + + ... +<br />
0! 1! 4!<br />
≈ 3.630170833<br />
7<br />
k<br />
1.3 1.3<br />
∑ k!<br />
k = 0<br />
= e ≈<br />
0 1 7<br />
1.3 1.3 1.3<br />
= + + ... +<br />
0! 1! 7!<br />
≈ 3.669060828<br />
<strong>12</strong>44<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
Section <strong>12</strong>.1: <strong>Sequences</strong><br />
97. a. a 1 = 0.4 ,<br />
2−2<br />
a 2 = 0.4 + 0.3⋅ 2 = 0.4 + 0.3 = 0.7<br />
3−2<br />
( )<br />
( )<br />
( )<br />
( )<br />
( )<br />
( )<br />
a 3<br />
= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 2 = 1.0<br />
4−2<br />
a 4<br />
= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 4 = 1.6<br />
5−2<br />
a 5 = 0.4 + 0.3⋅ 2 = 0.4 + 0.3 8 = 2.8<br />
6−2<br />
a 6<br />
= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 16 = 5.2<br />
7−2<br />
a 7<br />
= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 32 = 10.0<br />
8−2<br />
a 8 = 0.4 + 0.3⋅ 2 = 0.4 + 0.3 64 = 19.6<br />
The first eight terms of <strong>the</strong> sequence are 0.4,<br />
0.7, 1.0, 1.6, 2.8, 5.2, 10.0, and 19.6.<br />
b. Except for term 5, which has no match,<br />
Bode’s formula provides excellent<br />
approximations for <strong>the</strong> mean distances of<br />
<strong>the</strong> planets from <strong>the</strong> Sun.<br />
c. The mean distance of Ceres from <strong>the</strong> Sun is<br />
approximated by a 5 = 2.8 , and that of<br />
Uranus is a 8 = 19.6 .<br />
9−2<br />
d. = + ⋅ = + ( ) =<br />
10−2<br />
( )<br />
a 9<br />
0.4 0.3 2 0.4 0.3 <strong>12</strong>8 38.8<br />
a 10 = 0.4 + 0.3⋅ 2 = 0.4 + 0.3 256 = 77.2<br />
e. Pluto’s distance is approximated by a 9 , but<br />
no term approximates Neptune’s mean<br />
distance from <strong>the</strong> Sun.<br />
11−2<br />
f. ( )<br />
a 11<br />
= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 5<strong>12</strong> = 154<br />
According to Bode’s Law, <strong>the</strong> mean orbital<br />
distance of 2003 UB 313 will be 154 AU<br />
from <strong>the</strong> Sun.<br />
98. 5<br />
We begin with an initial guess of a<br />
0<br />
= 2 .<br />
1⎛<br />
5⎞<br />
a1<br />
= ⎜2+ ⎟=<br />
2.25<br />
2⎝<br />
2⎠<br />
1⎛<br />
5 ⎞<br />
a2<br />
= ⎜2.25 + ⎟≈2.236111111<br />
2⎝<br />
2.25⎠<br />
1⎛<br />
5 ⎞<br />
a3<br />
= ⎜2.236111111+<br />
⎟<br />
2 ⎝<br />
2.236111111⎠<br />
≈ 2.236067978<br />
1⎛<br />
5 ⎞<br />
a4<br />
= ⎜2.236067978<br />
+<br />
⎟<br />
2 ⎝<br />
2.236067978 ⎠<br />
≈ 2.236067977<br />
1⎛<br />
5 ⎞<br />
a5<br />
= ⎜2.236067977<br />
+<br />
⎟<br />
2 ⎝<br />
2.236067977 ⎠<br />
≈ 2.236067977<br />
For both a<br />
5<br />
and <strong>the</strong> calculator approximation,<br />
we obtain 5 ≈ 2.236067977 .<br />
99. 8<br />
We begin with an initial guess of a<br />
0<br />
= 3 .<br />
1⎛ 8 ⎞ 1⎛ 8⎞<br />
a1 = ⎜a0<br />
+ ⎟= ⎜3 + ⎟≈<br />
2.833333333<br />
2⎝<br />
a0<br />
⎠ 2⎝<br />
3⎠<br />
1⎛<br />
8 ⎞<br />
a2 = ⎜a1<br />
+ ⎟<br />
2 ⎝ a1<br />
⎠<br />
1⎛<br />
8 ⎞<br />
= ⎜2.833333333<br />
+<br />
⎟<br />
2 ⎝<br />
2.2.833333333 ⎠<br />
≈ 2.828431373<br />
1⎛<br />
8 ⎞<br />
a3 = ⎜a2<br />
+ ⎟<br />
2 ⎝ a2<br />
⎠<br />
1⎛<br />
8 ⎞<br />
= ⎜2.828431373<br />
+<br />
⎟<br />
2 ⎝<br />
2.828431373 ⎠<br />
≈ 2.828427<strong>12</strong>5<br />
1⎛<br />
8 ⎞<br />
a4 = ⎜a3<br />
+ ⎟<br />
2 ⎝ a3<br />
⎠<br />
1 ⎛<br />
8 ⎞<br />
= ⎜ 2.828427<strong>12</strong>5 +<br />
⎟<br />
2 ⎝<br />
2.828427<strong>12</strong>5 ⎠<br />
≈ 2.828427<strong>12</strong>5<br />
1⎛<br />
8 ⎞<br />
a5 = ⎜a4<br />
+ ⎟<br />
2 ⎝ a4<br />
⎠<br />
1⎛<br />
8 ⎞<br />
= ⎜2.828427<strong>12</strong>5<br />
+<br />
⎟<br />
2 ⎝<br />
2.828427<strong>12</strong>5 ⎠<br />
≈ 2.828427<strong>12</strong>5<br />
For both a<br />
5<br />
and <strong>the</strong> calculator approximation,<br />
we obtain 8 ≈ 2.828427<strong>12</strong>5 .<br />
<strong>12</strong>45<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
100. 21<br />
We begin with an initial guess of a<br />
0<br />
= 5 .<br />
1⎛<br />
21⎞<br />
a1<br />
= ⎜5+ ⎟=<br />
4.625<br />
2⎝<br />
5 ⎠<br />
1⎛<br />
21 ⎞<br />
a2<br />
= ⎜4.625 + ⎟≈<br />
4.58277027<br />
2⎝<br />
4.625⎠<br />
1⎛<br />
21 ⎞<br />
a3<br />
= ⎜4.58277027<br />
+<br />
⎟<br />
2 ⎝<br />
4.58277027 ⎠<br />
≈ 4.582575699<br />
1⎛<br />
21 ⎞<br />
a4<br />
= ⎜4.582575699<br />
+<br />
⎟<br />
2 ⎝<br />
4.582575699 ⎠<br />
≈ 4.582575695<br />
1⎛<br />
21 ⎞<br />
a5<br />
= ⎜4.582575695<br />
+<br />
⎟<br />
2 ⎝<br />
4.582575695 ⎠<br />
≈ 4.582575695<br />
101. 89<br />
We begin with an initial guess of a<br />
0<br />
= 9 .<br />
1⎛<br />
89⎞<br />
a1<br />
= ⎜5 + ⎟≈9.444444444<br />
2⎝<br />
5 ⎠<br />
1⎛<br />
89 ⎞<br />
a2<br />
= ⎜9.444444444<br />
+<br />
⎟<br />
2 ⎝<br />
9.444444444 ⎠<br />
≈ 9.433986928<br />
1⎛<br />
89 ⎞<br />
a3<br />
= ⎜9.433986928<br />
+<br />
⎟<br />
2 ⎝<br />
9.433986928 ⎠<br />
≈ 9.433981132<br />
1⎛<br />
89 ⎞<br />
a4<br />
= ⎜9.433981132<br />
+<br />
⎟<br />
2 ⎝<br />
9.433981132 ⎠<br />
≈ 9.433981132<br />
1⎛<br />
89 ⎞<br />
a5<br />
= ⎜9.433981132<br />
+<br />
⎟<br />
2 ⎝<br />
9.433981132<br />
⎠<br />
≈ 9.433981132<br />
For both a<br />
5<br />
and <strong>the</strong> calculator approximation,<br />
we obtain 21 ≈ 4.582575695 .<br />
For both a<br />
5<br />
and <strong>the</strong> calculator approximation,<br />
we obtain 89 ≈ 9.433981132 .<br />
_________________________________________________________________________________________________<br />
102. To show that 1 2 3 ... ( n 1)<br />
( + 1)<br />
n n<br />
+ + + + − + n =<br />
2<br />
S = 1 + 2 + 3 + ... + n− 1 + n, we can reverse <strong>the</strong> order to get<br />
Let ( )<br />
+ S = n+ ( n− ) + ( n−<br />
)<br />
1 2 +...+ 2 + 1, now add <strong>the</strong>se two lines to get<br />
[ ] ⎡ ( ) ⎤ ⎡ ( ) ⎤ ⎡( ) ⎤ [ ]<br />
2S = 1+ n + ⎣2 + n− 1 ⎦+ ⎣3+ n− 2 ⎦+ ...... + ⎣ n− 1 + 2⎦+ n+<br />
1<br />
So we have 2S = [ 1+ n] + [ 1+ n] + [ 1 + n] + .... + [ n+ 1] + [ n+ 1] = n⋅ [ n+<br />
1]<br />
Thus, 2S<br />
= n( n+<br />
1)<br />
n⋅ ( n+<br />
1)<br />
S =<br />
103. Answers will vary.<br />
2<br />
<strong>12</strong>46<br />
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Section <strong>12</strong>.2: Arithmetic <strong>Sequences</strong><br />
Section <strong>12</strong>.2<br />
1. arithmetic<br />
2. False; <strong>the</strong> sum of <strong>the</strong> first and last terms equals<br />
twice <strong>the</strong> sum of all <strong>the</strong> terms divided by <strong>the</strong><br />
number of terms.<br />
3. d = sn<br />
−sn−1<br />
= ( n+ 4) −( n− 1+ 4) = ( n+ 4) − ( n+<br />
3)<br />
= n+ 4−n− 3=<br />
1<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
s1 = 1+ 4= 5, s2 = 2+ 4= 6, s3<br />
= 3+ 4=<br />
7,<br />
s = 4+ 4=<br />
8<br />
4.<br />
5.<br />
6.<br />
4<br />
d = s −s<br />
n<br />
n−1<br />
( ) ( )<br />
= ( n−5) −( n−1− 5) = n−5 − n−6<br />
= n−5− n+ 6=<br />
1<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
s1 = 1− 5 =− 4, s2 = 2 − 5 =− 3, s3<br />
= 3 − 5 =−2,<br />
s = 4− 5= −1<br />
4<br />
d = an<br />
−an−1<br />
= ( 2n−5 ) −(2( n−1) −5)<br />
= ( 2n−5) −( 2n−2−5)<br />
= 2n−5− 2n+ 7 = 2<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
a1 = 21 ⋅ − 5= − 3, a2<br />
= 22 ⋅ − 5= −1,<br />
a = 23 ⋅ − 5= 1, a = 24 ⋅ − 5=<br />
3<br />
3 4<br />
d = bn<br />
−bn−1<br />
= ( 3n+ 1 ) −(3( n− 1) + 1)<br />
= ( 3n+ 1) −( 3n− 3+<br />
1)<br />
= 3n+ 1− 3n+ 2=<br />
3<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
b1 = 31 ⋅ + 1= 4, b2<br />
= 32 ⋅ + 1=<br />
7,<br />
b = 3⋅ 3+ 1= 10, b = 3⋅ 4+ 1=<br />
13<br />
3 4<br />
7.<br />
8.<br />
d = c − c<br />
n<br />
n−1<br />
( n)<br />
n<br />
( 6 2n) ( 6 2n<br />
2)<br />
= 6−2 −(6−2( −1))<br />
= − − − +<br />
= 6−2n− 6+ 2n− 2=−2<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
c = 6−2⋅ 1= 4, c = 6−2⋅ 2=<br />
2,<br />
c<br />
1 2<br />
= 6−2⋅ 3= 0, c = 6−2⋅ 4=−2<br />
3 4<br />
d = d − d<br />
n<br />
n−1<br />
( n)<br />
n<br />
( 4 2n) ( 4 2n<br />
2)<br />
= 4−2 −(4−2( −1))<br />
= − − − +<br />
= 4−2n− 4+ 2n− 2=−2<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
d = 4−2⋅ 1= 2, d = 4−2⋅ 2=<br />
0,<br />
d<br />
1 2<br />
= 4−2⋅ 3= − 2, d = 4−2⋅ 4= −4<br />
3 4<br />
9. d = tn<br />
− tn−1<br />
⎛1 1 ⎞ ⎛1 1 ⎞<br />
= ⎜ − n⎟−⎜ − ( n−<br />
1) ⎟<br />
⎝2 3 ⎠ ⎝2 3 ⎠<br />
⎛1 1 ⎞ ⎛1 1 1⎞<br />
= ⎜ − n⎟−⎜ − n+<br />
⎟<br />
⎝2 3 ⎠ ⎝2 3 3⎠<br />
1 1 1 1 1 1<br />
= − n− + n− = −<br />
2 3 2 3 3 3<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
1 1 1 1 1 1<br />
t1 = − ⋅ 1 = , t2<br />
= − ⋅ 2 =− ,<br />
2 3 6 2 3 6<br />
1 1 1 1 1 5<br />
t3 = − ⋅ 3 =− , t4<br />
= − ⋅ 4=−<br />
2 3 2 2 3 6<br />
10. d = tn<br />
− tn−1<br />
⎛2 1 ⎞ ⎛2 1 ⎞<br />
= ⎜ + n⎟− ⎜ + ( n−<br />
1) ⎟<br />
⎝3 4 ⎠ ⎝3 4 ⎠<br />
⎛2 1 ⎞ ⎛2 1 1⎞<br />
= ⎜ + n⎟− ⎜ + n−<br />
⎟<br />
⎝3 4 ⎠ ⎝3 4 4⎠<br />
2 1 2 1 1 1<br />
= + n− − n+ =<br />
3 4 3 4 4 4<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
2 1 11 2 1 7<br />
t1 = + ⋅ 1 = , t2<br />
= + ⋅ 2 = ,<br />
3 4 <strong>12</strong> 3 4 6<br />
2 1 17 2 1 5<br />
t3 = + ⋅ 3 = , t4<br />
= + ⋅ 4=<br />
3 4 <strong>12</strong> 3 4 3<br />
<strong>12</strong>47<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
11.<br />
d = s −s<br />
n<br />
n−1<br />
n n−1<br />
( ) ( )<br />
= ln 3 −ln 3<br />
nln ( 3) ( n 1) ln ( 3)<br />
( ln 3 )( n ( n 1 )) ( ln 3)( n n 1)<br />
= − −<br />
= − − = − +<br />
= ln 3<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
s<br />
1<br />
= ln 3 = ln 3 , s<br />
2<br />
= ln 3 = 2ln 3 ,<br />
s<br />
( ) ( ) ( ) ( )<br />
3 4<br />
( ) ( ) s ( ) ( )<br />
1 2<br />
= ln 3 = 3ln 3 , = ln 3 = 4ln 3<br />
3 4<br />
ln n ln( n−1)<br />
<strong>12</strong>. d s s e e n ( n )<br />
= n − n−1 = − = − − 1 = 1<br />
The difference between consecutive terms is<br />
constant, <strong>the</strong>refore <strong>the</strong> sequence is arithmetic.<br />
ln1 ln 2 ln 3<br />
1 2 3<br />
ln 4<br />
4 = e = 4<br />
s = e = 1, s = e = 2, s = e = 3,<br />
s<br />
13. an<br />
= a1 + ( n−1)<br />
d<br />
= 2 + ( n −1)3<br />
= 2+ 3n<br />
−3<br />
= 3n<br />
−1<br />
a 51 = 351 ⋅ − 1=<br />
152<br />
14. an<br />
= a1 + ( n−1)<br />
d<br />
=− 2 + ( n −1)4<br />
=− 2+ 4n<br />
−4<br />
= 4n<br />
−6<br />
a 51 = 451 ⋅ − 6=<br />
198<br />
15. an<br />
= a1 + ( n−1)<br />
d<br />
= 5 + ( n −1)( −3)<br />
= 5− 3n<br />
+ 3<br />
= 8−3n<br />
a 51 = 8−3⋅ 51=−<br />
145<br />
16. an<br />
= a1 + ( n−1)<br />
d<br />
= 6 + ( n −1)( −2)<br />
= 6− 2n<br />
+ 2<br />
= 8−2n<br />
a 51 = 8 −2⋅ 51 =− 94<br />
17.<br />
an<br />
= a1 + ( n−1)<br />
d<br />
1<br />
= 0 + ( n −1) 2<br />
1 1<br />
= n −<br />
2 2<br />
1<br />
= ( n −1)<br />
2<br />
1<br />
a 51 = ( 51 − 1 ) = 25<br />
2<br />
18. an<br />
= a1 + ( n−1)<br />
d<br />
⎛ 1 ⎞<br />
= 1 + ( n −1)<br />
⎜ − ⎟<br />
⎝ 3 ⎠<br />
1 1<br />
= 1− n + 3 3<br />
4 1<br />
= − n<br />
3 3<br />
4 1 4 51 47<br />
a 51 = − ⋅ 51 = − =−<br />
3 3 3 3 3<br />
19. an<br />
= a1 + ( n−1)<br />
d<br />
= 2 + ( n −1) 2<br />
= 2+ 2n− 2 = 2n<br />
a 51 = 51 2<br />
20. a a n d n ( n )<br />
n = 1 + ( − 1) = 0 + ( −1) π= −1<br />
π<br />
a 51 = 51π−π= 50π<br />
21. a1 = 2, d = 2, an<br />
= a1+ ( n−<br />
1) d<br />
a 100 = 2 + (100 − 1)2 = 2 + 99(2) = 2 + 198 = 200<br />
22. a1 = − 1, d = 2, an<br />
= a1+ ( n−<br />
1) d<br />
a 80 = − 1 + (80 − 1)2 = − 1+ 79(2) =− 1+ 158 = 157<br />
23. a1 = 1, d =−2 − 1 =− 3, an<br />
= a1+ ( n−<br />
1) d<br />
a 90 = 1 + (90−1)( − 3) = 1+ 89( −3)<br />
= 1− 267= −266<br />
24. a1 = 5, d = 0 − 5 =− 5, an<br />
= a1+ ( n−<br />
1) d<br />
a 80 = 5 + (80−1)( − 5) = 5+ 79( −5)<br />
= 5 − 395 = −390<br />
<strong>12</strong>48<br />
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Section <strong>12</strong>.2: Arithmetic <strong>Sequences</strong><br />
5 1<br />
25. a1 = 2, d = − 2 = , an<br />
= a1+ ( n−<br />
1) d<br />
2 2<br />
1 83<br />
a 80 = 2 + (80− 1) = 2 2<br />
26. a1<br />
= 2 5, d = 4 5− 2 5 = 2 5,<br />
an<br />
= a1<br />
+ ( n−1)<br />
d<br />
a 70 = 2 5 + (70−1)2 5<br />
= 2 5+<br />
69 2 5<br />
( )<br />
= 2 5+<br />
138 5<br />
= 140 5<br />
27. 8 1 20 1<br />
a = a + 7d = 8 a = a + 19d<br />
= 44<br />
Solve <strong>the</strong> system of equations by subtracting <strong>the</strong><br />
first equation from <strong>the</strong> second:<br />
<strong>12</strong>d<br />
= 36 ⇒ d = 3<br />
a1<br />
= 8− 7(3) = 8− 21=−13<br />
Recursive formula: a1 =− 13 an<br />
= an<br />
− 1+<br />
3<br />
nth term: an<br />
= a1 + ( n−1)<br />
d<br />
=− 13 + ( n −1)( 3)<br />
=− 13 + 3n<br />
−3<br />
= 3n<br />
−16<br />
28. 4 1 20 1<br />
a = a + 3d = 3 a = a + 19d<br />
= 35<br />
Solve <strong>the</strong> system of equations by subtracting <strong>the</strong><br />
first equation from <strong>the</strong> second:<br />
16d<br />
= 32 ⇒ d = 2<br />
a1<br />
= 3− 3(2) = 3− 6=−3<br />
Recursive formula: a1 =− 3 an<br />
= an<br />
− 1+<br />
2<br />
nth term: an<br />
= a1 + ( n−1)<br />
d<br />
=− 3+ ( n −1)( 2)<br />
=− 3+ 2n<br />
−2<br />
= 2n<br />
−5<br />
29. 9 1 15 1<br />
a = a + 8d = − 5 a = a + 14d<br />
= 31<br />
Solve <strong>the</strong> system of equations by subtracting <strong>the</strong><br />
first equation from <strong>the</strong> second:<br />
6d<br />
= 36⇒ d = 6<br />
a1<br />
=−5 − 8(6) =−5 − 48 =−53<br />
Recursive formula: a1 =− 53 an<br />
= an<br />
− 1+<br />
6<br />
nth term: an<br />
= a1 + ( n−1)<br />
d<br />
=− 53 + ( n −1)( 6)<br />
=− 53 + 6n<br />
−6<br />
= 6n<br />
−59<br />
30. 8 1 18 1<br />
a = a + 7d = 4 a = a + 17d<br />
=− 96<br />
Solve <strong>the</strong> system of equations by subtracting <strong>the</strong><br />
first equation from <strong>the</strong> second:<br />
10d<br />
= −100 ⇒ d =−10<br />
a = 4 −7( − 10) = 4 + 70 = 74<br />
1<br />
Recursive formula: a1 = 74 an<br />
= an<br />
− 1−<br />
10<br />
a = a + n−<br />
d<br />
nth term: n 1 ( 1)<br />
= 74 + ( n −1)( −10)<br />
= 74 − 10n<br />
+ 10<br />
= 84 −10n<br />
31. 15 1 40 1<br />
a = a + 14d = 0 a = a + 39d<br />
= − 50<br />
Solve <strong>the</strong> system of equations by subtracting <strong>the</strong><br />
first equation from <strong>the</strong> second:<br />
25d<br />
= −50 ⇒ d = −2<br />
a =−14( − 2) = 28<br />
1<br />
Recursive formula: a1 = 28 an<br />
= an<br />
− 1−<br />
2<br />
a = a + n−<br />
d<br />
nth term: n 1 ( 1)<br />
= 28 + ( n −1)( −2)<br />
= 28 − 2n<br />
+ 2<br />
= 30 −2n<br />
32. 5 1 13 1<br />
a = a + 4d =− 2 a = a + <strong>12</strong>d<br />
= 30<br />
Solve <strong>the</strong> system of equations by subtracting <strong>the</strong><br />
first equation from <strong>the</strong> second:<br />
8d<br />
= 32⇒ d = 4<br />
a = −2− 4(4) = −18<br />
1<br />
Recursive formula: a1 =− 18 an<br />
= an<br />
− 1+<br />
4<br />
a = a + n−<br />
d<br />
nth term: n 1 ( 1)<br />
=− 18 + ( n −1)( 4)<br />
=− 18 + 4n<br />
−4<br />
= 4n<br />
−22<br />
33. 14 1 18 1<br />
a = a + 13d = − 1 a = a + 17d<br />
=− 9<br />
Solve <strong>the</strong> system of equations by subtracting <strong>the</strong><br />
first equation from <strong>the</strong> second:<br />
4d<br />
= −8⇒ d = −2<br />
a = −1−13( − 2) = − 1+ 26=<br />
25<br />
1<br />
Recursive formula: a1 = 25 an<br />
= an<br />
− 1−<br />
2<br />
a = a + n−<br />
d<br />
nth term: n 1 ( 1)<br />
= 25 + ( n −1)( −2)<br />
= 25 − 2n<br />
+ 2<br />
= 27 −2n<br />
<strong>12</strong>49<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
34. <strong>12</strong> 1 18 1<br />
a = a + 11d = 4 a = a + 17d<br />
= 28<br />
Solve <strong>the</strong> system of equations by subtracting <strong>the</strong><br />
first equation from <strong>the</strong> second:<br />
6d<br />
= 24⇒ d = 4<br />
a = 4− 11(4) = 4− 44= −40<br />
1<br />
Recursive formula: a1 =− 40 an<br />
= an<br />
− 1+<br />
4<br />
a = a + n−<br />
d<br />
nth term: n 1 ( 1)<br />
=− 40 + ( n −1)( 4)<br />
=− 40 + 4n<br />
−4<br />
= 4n<br />
−44<br />
n n n<br />
1 n 1 2 1 2<br />
2 2 2<br />
2<br />
35. S = ( a + a ) = ( + ( n− )) = ( n) = n<br />
n<br />
n n<br />
1 n 2 2 1<br />
2 2<br />
2<br />
36. S = ( a + a ) = ( + n) = n+ n = n( n+<br />
)<br />
n<br />
n n n<br />
1 n 7 2 5 9 5<br />
2 2 2<br />
37. S = ( a + a ) = ( + ( + n)<br />
) = ( + n)<br />
n<br />
n n<br />
2<br />
1 n<br />
2<br />
n<br />
2<br />
= ( 4n− 6)<br />
= 2n −3n<br />
2<br />
= n −<br />
38. S = ( a + a ) = ( − 1+ ( 4n−5)<br />
)<br />
n<br />
( 2n<br />
3)<br />
39. a1 = 2, d = 4 − 2 = 2, an<br />
= a1+ ( n−<br />
1) d<br />
70 = 2 + ( n −1)2<br />
70 = 2 + 2n<br />
−2<br />
70 = 2n<br />
n = 35<br />
n 35 35<br />
Sn<br />
= ( a1<br />
+ an) = ( 2+ 70) = ( 72)<br />
= <strong>12</strong>60<br />
2 2 2<br />
40. a1 = 1, d = 3 − 1 = 2, an<br />
= a1+ ( n−<br />
1) d<br />
59 = 1 + ( n −1)2<br />
59 = 1+ 2n<br />
−2<br />
60 = 2n<br />
n = 30<br />
n 30<br />
Sn<br />
= ( a1<br />
+ an) = ( 1 + 59 ) = 15 ( 60 ) = 900<br />
2 2<br />
41. a1 = 5, d = 9 − 5 = 4, an<br />
= a1+ ( n−<br />
1) d<br />
49 = 5 + ( n −1)<br />
4<br />
49 = 5 + 4n<br />
−4<br />
48 = 4n<br />
n = <strong>12</strong><br />
n <strong>12</strong><br />
Sn<br />
= ( a1<br />
+ an) = ( 5 + 49 ) = 6 ( 54 ) = 324<br />
2 2<br />
42. a1 = 2, d = 5 − 2 = 3, an<br />
= a1+ ( n−<br />
1) d<br />
41 = 2 + ( n −1)<br />
3<br />
41 = 2 + 3n<br />
−3<br />
42 = 3n<br />
n = 14<br />
n 14<br />
Sn<br />
= ( a1<br />
+ an) = ( 2 + 41 ) = 7 ( 43 ) = 301<br />
2 2<br />
43. a 1 = 73 , 78 73 5<br />
d = − = ,<br />
n<br />
=<br />
1<br />
+ ( − 1)<br />
( n )( )<br />
( n )<br />
558 = 73 + −1 5<br />
485 = 5 −1<br />
97 = n −1<br />
98 = n<br />
n 98<br />
Sn<br />
= a1<br />
+ an<br />
= 73 + 558<br />
2 2<br />
= 49 631 = 30,919<br />
( ) ( )<br />
( )<br />
44. a 1 = 7 , 1 7 6<br />
a a n d<br />
d = − = − ,<br />
n = 1 + ( − 1)<br />
( n )( )<br />
( n )<br />
− 299 = 7 + −1 −6<br />
− 306 = −6 −1<br />
51 = n −1<br />
52 = n<br />
n 52<br />
Sn<br />
= a1<br />
+ an<br />
= 7 + − 299<br />
2 2<br />
= 26 − 292 = −7592<br />
a a n d<br />
( )<br />
( ) ( )<br />
( )<br />
45. a 1 = 4 , 4.5 4 0.5<br />
d = − = ,<br />
n<br />
=<br />
1<br />
+ ( − 1)<br />
( n )( )<br />
( n )<br />
100 = 4 + −1 0.5<br />
a a n d<br />
96 = 0.5 −1<br />
192 = n −1<br />
193 = n<br />
n 193 193<br />
Sn<br />
= a1<br />
+ an<br />
= 4+ 100 = 104<br />
2 2 2<br />
= 10,036<br />
( ) ( ) ( )<br />
<strong>12</strong>50<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
Section <strong>12</strong>.2: Arithmetic <strong>Sequences</strong><br />
1 1<br />
46. a 1 = 8 , d = 8 − 8= , an<br />
= a1 + ( n−<br />
1)<br />
d<br />
4 4<br />
⎛1<br />
⎞<br />
50 = 8 + ( n −1)<br />
⎜ ⎟<br />
⎝4<br />
⎠<br />
1<br />
42 = ( n −1)<br />
4<br />
168 = n −1<br />
169 = n<br />
n 169 169<br />
Sn<br />
= ( a1<br />
+ an) = ( 8 + 50) = ( 58)<br />
= 4901<br />
2 2 2<br />
47. a<br />
1<br />
= 21 () − 5= − 3, ( )<br />
a<br />
80<br />
= 280− 5=<br />
155<br />
80<br />
S<br />
80<br />
= ( − 3+ 155) = 40( 152)<br />
= 6080<br />
2<br />
48. a 1 = 3− 2()<br />
1 = 1, ( )<br />
a 90 = 3 − 2 90 =− 177<br />
( ( )) ( )<br />
90<br />
S<br />
90<br />
= 1 + − 177 = 45 − 176 = − 7920<br />
2<br />
1 11 1<br />
49. a<br />
1<br />
= 6− () 1 = , a<br />
100<br />
= 6 − ( 100)<br />
= − 44<br />
2 2<br />
2<br />
100 ⎛11 ⎞ ⎛ 77 ⎞<br />
S100<br />
= ⎜ + ( − 44)<br />
⎟= 50⎜− ⎟= −1925<br />
2 ⎝ 2 ⎠ ⎝ 2 ⎠<br />
1 1 5<br />
50. 1 () 1<br />
3 2 6<br />
1 1 163<br />
a = + =<br />
3 2 6<br />
a = + = , 80 ( 80)<br />
S<br />
80<br />
80 ⎛5 163 ⎞<br />
= ⎜ + ⎟= 40( 28)<br />
= 1<strong>12</strong>0<br />
2 ⎝6 6 ⎠<br />
d = − = , an<br />
= a1 + ( n−<br />
1)<br />
d<br />
a<br />
<strong>12</strong>0<br />
= 14 + ( <strong>12</strong>0 − 1)( 2) = 14 + 119( 2)<br />
= 252<br />
51. a 1 = 14 , 16 14 2<br />
<strong>12</strong>0<br />
S<br />
<strong>12</strong>0<br />
= ( 14 + 252 ) = 60 ( 266 ) = 15,960<br />
2<br />
d =− − =− , an<br />
= a1 + ( n−<br />
1)<br />
d<br />
a<br />
46<br />
= 2+ ( 46−1)( − 3) = 2+ ( 45)( − 3)<br />
= − 133<br />
52. a 1 = 2 , 1 2 3<br />
( ( )) ( )<br />
46<br />
S<br />
46<br />
= 2 + − 133 = 23 − 131 = − 3013<br />
2<br />
53. Find <strong>the</strong> common difference of <strong>the</strong> terms and<br />
solve <strong>the</strong> system of equations:<br />
(2x + 1) − ( x+ 3) = d ⇒ x− 2 = d<br />
(5x + 2) − (2x+ 1) = d ⇒ 3x+ 1 = d<br />
3x+ 1= x−2<br />
2x<br />
=−3<br />
3<br />
x =−<br />
2<br />
54. Find <strong>the</strong> common difference of <strong>the</strong> terms and<br />
solve <strong>the</strong> system of equations:<br />
(3x + 2) − (2 x) = d ⇒ x+ 2 = d<br />
(5x + 3) − (3x+ 2) = d ⇒ 2x+ 1 = d<br />
2x+ 1= x+<br />
2<br />
x = 1<br />
55. a 1 = 11 , d = 3 , S = 1092<br />
n<br />
( )<br />
an<br />
= a1 + ( n− 1) d = 11+ n− 1 3= 3n+<br />
8<br />
n<br />
Sn<br />
= ( a1<br />
+ an)<br />
2<br />
n<br />
1092 = ⎡11 (3 8)<br />
2<br />
⎣ + n + ⎤⎦<br />
n<br />
1092 = (3n<br />
+ 19)<br />
2<br />
2184 = n(3n+<br />
19)<br />
2<br />
0 = 3n<br />
+ 19n−2184<br />
0 = (3n+ 91)( n−24)<br />
91<br />
n =− or n = 24<br />
3<br />
Disregard <strong>the</strong> negative result. To obtain a sum<br />
of 1092, add 24 terms.<br />
56. a 1 = 78 , d = − 4 , S = 702<br />
n<br />
an<br />
= a1 + ( n− 1) d = 78 + ( n−1)( − 4) = − 4n+<br />
82<br />
n<br />
Sn<br />
= ( a1<br />
+ an)<br />
2<br />
n<br />
702 = ⎡78 ( 4 82)<br />
2<br />
⎣ + − n + ⎤⎦<br />
n<br />
702 = ( − 4n<br />
+ 160)<br />
2<br />
2<br />
0 =− 2n<br />
+ 80n−702<br />
2<br />
0= n − 40n+<br />
351<br />
0 = ( n−27)( n−13)<br />
n= 27 or n=<br />
13<br />
To obtain a sum of 702, add 13 terms or 27 terms.<br />
<strong>12</strong>51<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
57. The total number of seats is:<br />
S = 25 + 26 + 27 + + ( 25 + 29()<br />
1 )<br />
This is <strong>the</strong> sum of an arithmetic sequence with<br />
d = 1, a1<br />
= 25, and n= 30 .<br />
Find <strong>the</strong> sum of <strong>the</strong> sequence:<br />
30<br />
S 30 = [ 2(25) + (30 − 1)(1) ]<br />
2<br />
= 15(50 + 29) = 15(79)<br />
= 1185<br />
There are 1185 seats in <strong>the</strong> <strong>the</strong>ater.<br />
58. The total number of seats is:<br />
( ( ( )))<br />
S = 15 + 17 + 19 + + 15 + 39 2<br />
This is <strong>the</strong> sum of an arithmetic sequence with<br />
d = 2, a1<br />
= 15, and n = 40 .<br />
Find <strong>the</strong> sum of <strong>the</strong> sequence:<br />
40<br />
S 40 = [ 2(15) + (40 − 1)(2) ]<br />
2<br />
= 20(30 + 78) = 20(108)<br />
= 2160<br />
The corner section has 2160 seats.<br />
59. The lighter colored tiles have 20 tiles in <strong>the</strong><br />
bottom row and 1 tile in <strong>the</strong> top row. The<br />
number decreases by 1 as we move up <strong>the</strong><br />
triangle. This is an arithmetic sequence with<br />
a1 = 20, d =− 1, and n = 20 . Find <strong>the</strong> sum:<br />
20<br />
S = [ 2(20) + (20 − 1)( − 1) ]<br />
2<br />
= 10(40 − 19) = 10(21)<br />
= 210<br />
There are 210 lighter tiles.<br />
The darker colored tiles have 19 tiles in <strong>the</strong><br />
bottom row and 1 tile in <strong>the</strong> top row. The<br />
number decreases by 1 as we move up <strong>the</strong><br />
triangle. This is an arithmetic sequence with<br />
a1 = 19, d =− 1, and n = 19 . Find <strong>the</strong> sum:<br />
19<br />
S = [ 2(19) + (19 − 1)( − 1) ]<br />
2<br />
19 19<br />
= (38 − 18) = (20) = 190<br />
2 2<br />
There are 190 darker tiles.<br />
60. The number of bricks required decreases by 2 on<br />
each successive step. This is an arithmetic<br />
sequence with a1 = 100, d =− 2, and n= 30 .<br />
a. The number of bricks for <strong>the</strong> top step is:<br />
a30 = a1 + ( n− 1) d = 100 + (30 −1)( −2)<br />
= 100 + 29( − 2) = 100 −58<br />
= 42<br />
42 bricks are required for <strong>the</strong> top step.<br />
b. The total number of bricks required is <strong>the</strong><br />
sum of <strong>the</strong> sequence:<br />
30<br />
S = [ 100 + 42 ] = 15(142) = 2130<br />
2<br />
2130 bricks are required to build <strong>the</strong><br />
staircase.<br />
61. The air cools at <strong>the</strong> rate of 5.5° F per 1000 feet.<br />
Since n represents thousands of feet, we have<br />
d = − 5.5 . The ground temperature is 67° F so<br />
we have T 1 = 67 − 5.5 = 61.5 . Therefore,<br />
T = 61.5 + n−1 −5.5<br />
{ }<br />
{ n} ( )( )<br />
= { − 5.5n+ 67 } or { 67 −5.5n}<br />
After <strong>the</strong> parcel of air has risen 5000 feet, we<br />
have T 5 = 61.5 + ( 5 −1)( − 5.5)<br />
= 39.5 .<br />
The parcel of air will be 39.5° F after it has risen<br />
5000 feet.<br />
62. If we treat <strong>the</strong> length of each rung as <strong>the</strong> term of<br />
an arithmetic sequence, we have 1 49 a = ,<br />
d = − 2.5 , and a = 24 .<br />
n<br />
n<br />
( )<br />
( n )( )<br />
( n )<br />
a = a + n−<br />
d<br />
1<br />
1<br />
24 = 49 + −1 −2.5<br />
− 25 =−2.5 −1<br />
10 = n −1<br />
11 = n<br />
Therefore, <strong>the</strong> ladder contains 11 rungs.<br />
To find <strong>the</strong> total material required for <strong>the</strong> rungs,<br />
we need <strong>the</strong> sum of <strong>the</strong>ir lengths. Since <strong>the</strong>re are<br />
11 rungs, we have<br />
11 11<br />
S<br />
11<br />
= ( 49 + 24) = ( 73)<br />
= 401.5<br />
2 2<br />
It would require 401.5 feet of material to<br />
construct <strong>the</strong> rungs for <strong>the</strong> ladder.<br />
<strong>12</strong>52<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
Section <strong>12</strong>.3: Geometric <strong>Sequences</strong>; Geometric Series<br />
63. a 1 = 35 , 37 35 2<br />
d = − = ,<br />
n<br />
=<br />
1<br />
+ ( − 1)<br />
a 27 = 35 + ( 27 − 1)( 2) = 35 + 26( 2)<br />
= 87<br />
a a n d<br />
27 27<br />
S<br />
27<br />
= ( 35 + 87) = ( <strong>12</strong>2)<br />
= 1647<br />
2 2<br />
The amphi<strong>the</strong>ater has 1647 seats.<br />
64. Find n in an arithmetic sequence with<br />
a1 = 10, d = 4, S n = 2040 .<br />
n<br />
Sn<br />
= [ 2 a1<br />
+ ( n−1)<br />
d]<br />
2<br />
n<br />
2040 = [ 2(10) + ( n −1)4]<br />
2<br />
4080 = n[ 20 + 4n−4]<br />
4080 = n(4n+<br />
16)<br />
2<br />
4080 = 4n<br />
+ 16n<br />
2<br />
1020 = n + 4n<br />
2<br />
n + 4n− 1020=<br />
0<br />
( n+ 34)( n− 30) = 0 ⇒ n =− 34 or n=<br />
30<br />
There are 30 rows in <strong>the</strong> corner section of <strong>the</strong><br />
stadium.<br />
65. The yearly salaries form an arithmetic sequence<br />
with a1 = 35,000, d = 1400, S n = 280,000 .<br />
Find <strong>the</strong> number of years for <strong>the</strong> aggregate salary<br />
to equal $280,000.<br />
n<br />
Sn<br />
= [ 2 a1<br />
+ ( n−1)<br />
d]<br />
2<br />
n<br />
280,000 = [ 2(35,000) + ( n −1)1400]<br />
2<br />
280,000 = n[ 35,000 + 700n−700]<br />
280,000 = n(700n+<br />
34,300)<br />
2<br />
280,000 = 700 n + 34,300 n<br />
2<br />
400 = n + 49 n<br />
2<br />
n + 49 n− 400 = 0<br />
2<br />
− 49 ± 49 −4(1)( −400)<br />
n =<br />
2(1)<br />
− 49 ± 4001 − 49 ± 63.25<br />
= ≈<br />
2 2<br />
n≈7.13 or n ≈− 56.13<br />
It takes about 8 years to have an aggregate salary<br />
of at least $280,000. The aggregate salary after 8<br />
years will be $319,200.<br />
66. Answers will vary.<br />
67. Answers will vary. Both increase (or decrease) at<br />
a constant rate, but <strong>the</strong> domain of an arithmetic<br />
sequence is <strong>the</strong> set of natural numbers while <strong>the</strong><br />
domain of a linear function is <strong>the</strong> set of all real<br />
numbers.<br />
Section <strong>12</strong>.3<br />
1. geometric<br />
2.<br />
a<br />
1− r<br />
3. divergent series<br />
4. True<br />
5. False; <strong>the</strong> common ratio can be positive or<br />
negative (or 0, but this results in a sequence of<br />
only 0s).<br />
6. True<br />
7.<br />
8.<br />
n+<br />
1<br />
3 n+−<br />
1 n<br />
r = = 3 = 3<br />
n<br />
3<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
1 2<br />
1 = = s2<br />
= =<br />
3 4<br />
3 s4<br />
s<br />
s<br />
3 3, 3 9,<br />
= 3 = 27, = 3 = 81<br />
n+<br />
1<br />
( −5)<br />
n+−<br />
1 n<br />
r = = ( − 5) = −5<br />
n<br />
( −5)<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
1 2<br />
1 = − =− s2<br />
= − =<br />
3 4<br />
3 s4<br />
s<br />
s<br />
( 5) 5, ( 5) 25,<br />
= ( − 5) = − <strong>12</strong>5, = ( − 5) = 625<br />
n+<br />
1<br />
⎛1<br />
⎞<br />
−3<br />
⎜ ⎟ n+−<br />
1 n<br />
2 ⎛1⎞<br />
1<br />
9. r =<br />
⎝ ⎠<br />
=<br />
n ⎜ ⎟ =<br />
⎛1<br />
⎞ ⎝2⎠<br />
2<br />
−3 ⎜ ⎟<br />
⎝ 2 ⎠<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
1 2<br />
⎛1⎞ 3 ⎛1⎞<br />
3<br />
a1 =− 3 ⎜ ⎟ =− , a2<br />
=− 3 ⎜ ⎟ =− ,<br />
⎝2⎠ 2 ⎝2⎠<br />
4<br />
3 4<br />
⎛1⎞ 3 ⎛1⎞<br />
3<br />
a3 = − 3 ⎜ ⎟ = − , a4<br />
=− 3⎜ ⎟ =−<br />
⎝2⎠ 8 ⎝2⎠<br />
16<br />
<strong>12</strong>53<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
10.<br />
11.<br />
<strong>12</strong>.<br />
13.<br />
n+<br />
1<br />
⎛5<br />
⎞<br />
⎜ ⎟<br />
n+−<br />
1 n<br />
2 ⎛5⎞<br />
5<br />
r =<br />
⎝ ⎠<br />
=<br />
n ⎜ ⎟ =<br />
⎛5<br />
⎞ ⎝2⎠<br />
2<br />
⎜ ⎟<br />
⎝2<br />
⎠<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
1 2<br />
⎛5⎞ 5 ⎛5⎞<br />
25<br />
b1 = ⎜ ⎟ = , b2<br />
= ⎜ ⎟ = ,<br />
⎝2⎠ 2 ⎝2⎠<br />
4<br />
3 4<br />
⎛5 ⎞ <strong>12</strong>5 ⎛5 ⎞ 625<br />
b3 = ⎜ ⎟ = , b4<br />
= ⎜ ⎟ =<br />
⎝2⎠ 8 ⎝2⎠<br />
16<br />
⎛<br />
n+−<br />
11<br />
2 ⎞<br />
⎜<br />
4 ⎟ n<br />
2 n−( n−1)<br />
r =<br />
⎝ ⎠<br />
= = 2 = 2<br />
n−1<br />
n−1<br />
⎛2<br />
⎞ 2<br />
⎜<br />
4 ⎟<br />
⎝ ⎠<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
11 − 0<br />
2 2 −2<br />
1<br />
c1 = = = 2 = ,<br />
2<br />
4 2 4<br />
2−1 1<br />
2 2 −1<br />
1<br />
c2 = = = 2 = ,<br />
2<br />
4 2 2<br />
3−1 2<br />
2 2<br />
3 2<br />
c<br />
c<br />
= = = 1,<br />
4 2<br />
4−1 3<br />
2 2<br />
4 2<br />
= = = 2<br />
4 2<br />
⎛<br />
n+<br />
1<br />
3 ⎞<br />
⎜<br />
9 ⎟ n+<br />
1<br />
3 n+−<br />
1 n<br />
r =<br />
⎝ ⎠<br />
= = 3 = 3<br />
n n<br />
⎛3<br />
⎞ 3<br />
⎜<br />
9 ⎟<br />
⎝ ⎠<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
1 2<br />
1 d2<br />
3 1 3 9<br />
d = = , = = = 1,<br />
9 3 9 9<br />
3 4<br />
3 27 3 81<br />
d3 = = = 3, d4<br />
= = = 9<br />
9 9 9 9<br />
⎛n+<br />
1⎞<br />
⎜ ⎟<br />
3<br />
⎛n+<br />
1 n⎞<br />
⎝ ⎠ ⎜ − ⎟<br />
⎝ 3 3⎠<br />
⎛n<br />
⎞<br />
⎜ ⎟<br />
⎝3<br />
⎠<br />
2<br />
r = = 2 = 2<br />
1/3<br />
2<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
1/3 2/3 3/3 4/3<br />
1 = 2 , 2 = 2 , 3 = 2 = 2, 4 = 2<br />
e e e e<br />
14.<br />
15.<br />
16.<br />
17.<br />
18.<br />
19.<br />
20.<br />
2( n+<br />
1)<br />
3<br />
2n+ 2−2n<br />
2<br />
r = = 3 = 3 = 9<br />
2n<br />
3<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
21 ⋅<br />
22 ⋅ 4<br />
f1 = 3 = 9, f2<br />
= 3 = 3 = 81,<br />
23 ⋅ 6 24 ⋅ 8<br />
f = 3 = 3 = 729, f = 3 = 3 = 6561<br />
3 4<br />
⎛<br />
n+−<br />
11<br />
3 ⎞<br />
⎜ n+<br />
1<br />
2 ⎟ n n<br />
3 2<br />
r =<br />
⎝ ⎠<br />
= ⋅<br />
n 1 n− 1 n+<br />
1<br />
⎛<br />
−<br />
3 ⎞ 3 2<br />
⎜ n<br />
2 ⎟<br />
⎝ ⎠<br />
n−( n−1) n− ( n+ 1) −1<br />
3<br />
= 3 ⋅ 2 = 3⋅ 2 =<br />
2<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
11 − 0 21 − 1<br />
1 t<br />
1 2 2 2<br />
3 3 1 3 3 3<br />
t = = = , = = = ,<br />
2 2 2 2 2 4<br />
3−1 2 4−1 3<br />
3 3 9 3 3 27<br />
t3 = = = , t<br />
3 3 4 = = =<br />
4 4<br />
2 2 8 2 2 16<br />
⎛<br />
n+<br />
1<br />
2 ⎞<br />
⎜ n+− 11 1 1<br />
3 ⎟ n− n+<br />
3 2<br />
r =<br />
⎝ ⎠<br />
= ⋅<br />
n n n<br />
⎛ 2 ⎞ 3 2<br />
⎜ n−1<br />
3 ⎟<br />
⎝ ⎠<br />
n−− 1 n n+− 1 n −1<br />
2<br />
= 3 ⋅ 2 = 3 ⋅ 2=<br />
3<br />
The ratio of consecutive terms is constant,<br />
<strong>the</strong>refore <strong>the</strong> sequence is geometric.<br />
1 2<br />
1 11 − 0<br />
u2<br />
21 −<br />
2 2 2 2 4<br />
u = = = = 2, = = ,<br />
3 3 1 3 3<br />
3 4<br />
2 8 8 2 16 16<br />
u3 = = = , u<br />
3−1 2 4 = = =<br />
4−1 3<br />
3 3 9 3 3 27<br />
a<br />
a<br />
5<br />
n<br />
a<br />
a<br />
5<br />
n<br />
a<br />
a<br />
5<br />
n<br />
n<br />
5−1 4<br />
= 23 ⋅ = 23 ⋅ = 281 ⋅ = 162<br />
1<br />
23 n −<br />
= ⋅<br />
5−1 4<br />
=−2⋅ 4 =−2⋅ 4 =−2⋅ 256 =−5<strong>12</strong><br />
1<br />
24 n −<br />
=− ⋅<br />
5−1 4<br />
= 5( − 1) = 5( − 1) = 5⋅ 1 = 5<br />
1<br />
5(1) n −<br />
= ⋅ −<br />
5−1 4<br />
a5<br />
= 6( − 2) = 6( − 2) = 6⋅ 16 = 96<br />
1<br />
a 6( 2) n −<br />
= ⋅ −<br />
<strong>12</strong>54<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
Section <strong>12</strong>.3: Geometric <strong>Sequences</strong>; Geometric Series<br />
21.<br />
22.<br />
a<br />
a<br />
a<br />
a<br />
5<br />
n<br />
5<br />
n<br />
5−1 4<br />
⎛1⎞ ⎛1⎞<br />
= 0⋅ ⎜ ⎟ = 0⋅ ⎜ ⎟ = 0<br />
⎝2⎠ ⎝2⎠<br />
n−1<br />
⎛1<br />
⎞<br />
= 0⋅ ⎜ ⎟ = 0<br />
⎝2<br />
⎠<br />
5−1 4<br />
⎛ 1⎞ ⎛ 1⎞<br />
1<br />
= 1⋅⎜− ⎟ = 1⋅⎜− ⎟ =<br />
⎝ 3⎠ ⎝ 3⎠<br />
81<br />
n−1 n−1<br />
⎛ 1⎞ ⎛ 1⎞<br />
= 1⋅⎜− ⎟ = ⎜−<br />
⎟<br />
⎝ 3⎠ ⎝ 3⎠<br />
−<br />
23. a5<br />
( ) ( )<br />
n−1<br />
n<br />
an<br />
= 2⋅ ( 2) = ( 2)<br />
24.<br />
a<br />
a<br />
5<br />
n<br />
5 1 4<br />
= 2⋅ 2 = 2⋅ 2 = 2⋅ 4=<br />
4 2<br />
5−1 4<br />
⎛1⎞ ⎛1⎞<br />
= 0⋅ ⎜ ⎟ = 0⋅ ⎜ ⎟ = 0<br />
⎝π⎠ ⎝π⎠<br />
n−1<br />
⎛1<br />
⎞<br />
= 0⋅ ⎜ ⎟ = 0<br />
⎝π<br />
⎠<br />
1<br />
25. a1<br />
= 1, r = , n = 7<br />
2<br />
a<br />
7<br />
7−1 6<br />
⎛1⎞ ⎛1⎞<br />
1<br />
= 1⋅ ⎜ ⎟ = ⎜ ⎟ =<br />
⎝ 2 ⎠ ⎝ 2 ⎠ 64<br />
26. a1<br />
= 1, r = 3, n = 8<br />
27.<br />
a<br />
8<br />
1<br />
9<br />
8−1 7<br />
= 1⋅ 3 = 3 = 2187<br />
a = 1, r =− 1, n = 9<br />
a<br />
−<br />
( ) ( )<br />
9 1 8<br />
= 1⋅ − 1 = − 1 = 1<br />
10<br />
n<br />
32. a 1 = 5 , r = = 2 , an<br />
= a1<br />
⋅ r −<br />
5<br />
1<br />
52 n −<br />
= ⋅<br />
a n<br />
1 1<br />
n<br />
33. a 1 = − 3 , r = =− , an<br />
= a1<br />
⋅ r −<br />
−3 3<br />
n−1 n−2<br />
⎛ 1⎞ ⎛ 1⎞<br />
a n =−3⎜− ⎟ = ⎜−<br />
⎟<br />
⎝ 3⎠ ⎝ 3⎠<br />
1<br />
n<br />
34. a 1 = 4 , r = , an<br />
= a1<br />
⋅ r −<br />
4<br />
n−1 n−2<br />
⎛1⎞ ⎛1⎞<br />
a n = 4⎜ ⎟ = ⎜ ⎟<br />
⎝ 4 ⎠ ⎝ 4 ⎠<br />
35.<br />
36.<br />
n 1<br />
an<br />
= a1<br />
⋅ r −<br />
243 = a ⋅ −3<br />
1<br />
1<br />
( )<br />
5<br />
( )<br />
243 = a1<br />
−3<br />
243 =−243a<br />
− 1 = a<br />
Therefore,<br />
a = a ⋅ r −<br />
1<br />
a n<br />
n 1<br />
n 1<br />
2−1<br />
⎛1<br />
⎞<br />
7 = a1<br />
⎜ ⎟<br />
⎝3<br />
⎠<br />
1<br />
7 = a1<br />
3<br />
21 = a<br />
1<br />
Therefore,<br />
a n<br />
6−1<br />
=−− ( 3) n<br />
−1<br />
n<br />
⎛1<br />
⎞<br />
= 21⎜ ⎟<br />
⎝ 3 ⎠<br />
−1<br />
.<br />
1<br />
1<br />
1<br />
28.<br />
29.<br />
30.<br />
a =− 1, r =− 2, n=<br />
10<br />
a<br />
1<br />
10<br />
1<br />
8<br />
−<br />
( ) ( )<br />
10 1 9<br />
=−1⋅ − 2 = −1⋅ − 2 =−1( − 5<strong>12</strong>) = 5<strong>12</strong><br />
a = 0.4, r = 0.1, n=<br />
8<br />
a<br />
1<br />
7<br />
−<br />
( ) ( )<br />
8 1 7<br />
= 0.4⋅ 0.1 = 0.4 0.1 = 0.00000004<br />
a = 0.1, r = 10, n=<br />
7<br />
a<br />
7−1<br />
( )<br />
= 0.1⋅ 10 = 0.1 10 = 100,000<br />
14<br />
n<br />
31. a 1 = 7 , r = = 2 , an<br />
= a1<br />
⋅ r −<br />
7<br />
1<br />
7 2 n −<br />
= ⋅<br />
a n<br />
6<br />
1<br />
37.<br />
2 1<br />
4−1 3<br />
a4 a1⋅<br />
r r<br />
= = = r<br />
2−1<br />
a a ⋅ r r<br />
2 1575<br />
r = = 225<br />
7<br />
r = 225 = 15<br />
a = a ⋅r<br />
n<br />
1<br />
1<br />
1<br />
n−1<br />
7 = a ⋅15<br />
7 = 15a<br />
7<br />
a1<br />
=<br />
15<br />
Therefore,<br />
2−1<br />
a n<br />
2<br />
7 15<br />
n−1 7 15<br />
n−2<br />
= ⋅ = ⋅ .<br />
15<br />
<strong>12</strong>55<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
38.<br />
39.<br />
3 1<br />
6−1 5<br />
a6 a1<br />
⋅ r r<br />
= = = r<br />
3−1 2<br />
a a ⋅ r r<br />
r<br />
1<br />
3 81<br />
1<br />
3<br />
1 1<br />
= = ⋅ 3 =<br />
81 27<br />
1 1<br />
r = 3 =<br />
27 3<br />
n−1<br />
a = a ⋅r<br />
n 1<br />
3−1<br />
1 ⎛1⎞<br />
= a1<br />
⋅ ⎜ ⎟<br />
3 ⎝3⎠<br />
1 1<br />
= a1<br />
3 9<br />
3 = a<br />
1<br />
Therefore,<br />
a n<br />
3<br />
n−1 n−2<br />
⎛1⎞ ⎛1⎞<br />
= 3⎜ ⎟ = ⎜ ⎟<br />
⎝ 3 ⎠ ⎝ 3 ⎠<br />
1<br />
a1<br />
= , r = 2<br />
4<br />
⎛<br />
n<br />
n<br />
1−r<br />
⎞ 1⎛1−2 ⎞ 1<br />
Sn<br />
= a1<br />
⎜ 1 2<br />
1 r ⎟<br />
= 4⎜ 1 2 ⎟<br />
= − −<br />
⎝ − ⎠ ⎝ − ⎠ 4<br />
1<br />
( 2<br />
n<br />
= − 1 )<br />
4<br />
.<br />
n<br />
( )<br />
42.<br />
a = 4, r = 3<br />
S<br />
1<br />
n<br />
⎛<br />
n n n<br />
1−r<br />
⎞ ⎛1−3 ⎞ ⎛1−3<br />
⎞<br />
= a1<br />
⎜ = 4 = 4<br />
1−r<br />
⎟ ⎜ 1−3 ⎟ ⎜ −2<br />
⎟<br />
⎝ ⎠ ⎝ ⎠ ⎝ ⎠<br />
n n<br />
( ) ( )<br />
=−21− 3 = 23 −1<br />
43. a1<br />
= − 1, r = 2<br />
S<br />
n<br />
⎛<br />
n<br />
n<br />
1−r<br />
⎞ ⎛1−2<br />
⎞<br />
= a1<br />
⎜ =− 1 = 1−2<br />
1−r<br />
⎟ ⎜ 1−2<br />
⎟<br />
⎝ ⎠ ⎝ ⎠<br />
3<br />
44. a1<br />
= 2, r =<br />
5<br />
S<br />
n<br />
⎡<br />
n<br />
n<br />
⎛3⎞ ⎤ ⎡ ⎛3⎞<br />
⎤<br />
n ⎢1−<br />
⎥ ⎢1−<br />
⎥<br />
⎛1− r ⎞ ⎜ ⎟ ⎜ ⎟<br />
5 5<br />
= a1<br />
2<br />
⎝ ⎠<br />
2<br />
⎝ ⎠<br />
⎢ ⎥ ⎢ ⎥<br />
⎜ 1 r ⎟<br />
= =<br />
⎢ 3 ⎥ ⎢ 2 ⎥<br />
⎝ − ⎠ 1<br />
⎛ ⎞<br />
⎢ − 5<br />
⎥ ⎢ ⎜ ⎟ ⎥<br />
⎢⎣ ⎥⎦ ⎢⎣ ⎝ 5 ⎠ ⎥⎦<br />
⎡<br />
n<br />
⎛3<br />
⎞ ⎤<br />
= 51 ⎢ −⎜<br />
⎟ ⎥<br />
⎢⎣<br />
⎝5<br />
⎠ ⎥⎦<br />
45. Using <strong>the</strong> sum of <strong>the</strong> sequence feature:<br />
n<br />
40.<br />
3 1<br />
a1<br />
= = , r = 3<br />
9 3<br />
⎛<br />
n n n<br />
1−r<br />
⎞ 1⎛1−3 ⎞ 1⎛1−3<br />
⎞<br />
Sn<br />
= a1<br />
⎜ 1 r ⎟<br />
= 3⎜ 1 3 ⎟<br />
=<br />
3⎜ 2 ⎟<br />
⎝ − ⎠ ⎝ − ⎠ ⎝ − ⎠<br />
1 n 1 n<br />
=− ( 1− 3 ) = ( 3 −1)<br />
6 6<br />
46. Using <strong>the</strong> sum of <strong>the</strong> sequence feature:<br />
2 2<br />
41. a1<br />
= , r =<br />
3 3<br />
S<br />
n<br />
⎡<br />
n<br />
⎛2<br />
⎞ ⎤<br />
n ⎢1−<br />
⎥<br />
⎛1− r ⎞ 2<br />
⎜ ⎟<br />
3<br />
= a<br />
⎝ ⎠<br />
1<br />
⎢ ⎥<br />
⎜ 1 r ⎟<br />
=<br />
3⎢ 2 ⎥<br />
⎝ − ⎠<br />
⎢ 1−<br />
3<br />
⎥<br />
⎢⎣<br />
⎥⎦<br />
⎡<br />
n<br />
⎛2<br />
⎞ ⎤<br />
⎢1−<br />
2<br />
⎜ ⎟ ⎥<br />
n<br />
3 ⎡ ⎛2⎞<br />
⎤<br />
= ⎢ ⎝ ⎠ ⎥ = 21 ⎢ − ⎥<br />
3⎢ 1 ⎥ ⎜ ⎟<br />
⎢ ⎝3⎠<br />
⎥<br />
⎢<br />
⎣ ⎦<br />
3<br />
⎥<br />
⎢⎣<br />
⎥⎦<br />
47. Using <strong>the</strong> sum of <strong>the</strong> sequence feature:<br />
48. Using <strong>the</strong> sum of <strong>the</strong> sequence feature:<br />
<strong>12</strong>56<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
Section <strong>12</strong>.3: Geometric <strong>Sequences</strong>; Geometric Series<br />
49. Using <strong>the</strong> sum of <strong>the</strong> sequence feature:<br />
50. Using <strong>the</strong> sum of <strong>the</strong> sequence feature:<br />
1<br />
51. a1<br />
= 1, r =<br />
3<br />
Since r < 1, <strong>the</strong> series converges.<br />
S<br />
∞<br />
a1<br />
1 1 3<br />
= = = =<br />
1− r ⎛ 1⎞ ⎛2⎞<br />
2<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 3 ⎠ ⎝ 3 ⎠<br />
2<br />
52. a1<br />
= 2, r =<br />
3<br />
Since r < 1, <strong>the</strong> series converges.<br />
S<br />
∞<br />
a1<br />
2 2<br />
= = = = 6<br />
1− r ⎛ 2⎞ ⎛1⎞<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 3 ⎠ ⎝ 3 ⎠<br />
1<br />
53. a1<br />
= 8, r =<br />
2<br />
Since r < 1, <strong>the</strong> series converges.<br />
S<br />
∞<br />
a1<br />
8 8<br />
= = = = 16<br />
1− r ⎛ 1⎞ ⎛1⎞<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 2 ⎠ ⎝ 2 ⎠<br />
1<br />
54. a1<br />
= 6, r =<br />
3<br />
Since r < 1, <strong>the</strong> series converges.<br />
S<br />
∞<br />
a1<br />
6 6<br />
= = = = 9<br />
1− r ⎛ 1⎞ ⎛2⎞<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 3 ⎠ ⎝ 3 ⎠<br />
1<br />
55. a1<br />
= 2, r =−<br />
4<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
2 2 8<br />
S∞<br />
= = = =<br />
1− r ⎛ ⎛ 1 ⎞⎞ ⎛5<br />
⎞ 5<br />
⎜1−⎜− ⎟ ⎜ ⎟<br />
4<br />
⎟<br />
⎝ ⎝ ⎠⎠<br />
⎝4<br />
⎠<br />
3<br />
56. a1<br />
= 1, r =−<br />
4<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
1 1 4<br />
S∞<br />
= = = =<br />
1− r ⎛ ⎛ 3 ⎞⎞ ⎛7<br />
⎞ 7<br />
⎜1−⎜− ⎟ ⎜ ⎟<br />
4<br />
⎟<br />
⎝ ⎝ ⎠⎠<br />
⎝4<br />
⎠<br />
57. a 1 = 8 ,<br />
3<br />
r =<br />
2<br />
Since r > 1, <strong>the</strong> series diverges.<br />
58. a 1 = 9 ,<br />
4<br />
r =<br />
3<br />
Since r > 1, <strong>the</strong> series diverges.<br />
1<br />
59. a1<br />
= 5, r =<br />
4<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
5 5 20<br />
S∞<br />
= = = =<br />
1− r ⎛ 1⎞ ⎛3⎞<br />
3<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 4 ⎠ ⎝ 4 ⎠<br />
1<br />
60. a1<br />
= 8, r =<br />
3<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
8 8<br />
S∞<br />
= = = = <strong>12</strong><br />
1− r ⎛ 1⎞ ⎛2⎞<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 3 ⎠ ⎝ 3 ⎠<br />
1<br />
61. a<br />
1<br />
= , r = 3<br />
2<br />
Since r > 1, <strong>the</strong> series diverges.<br />
62. a 1 = 3 ,<br />
3<br />
r =<br />
2<br />
Since r > 1, <strong>the</strong> series diverges.<br />
<strong>12</strong>57<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
2<br />
63. a1<br />
= 6, r = −<br />
3<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
6 6 18<br />
S∞<br />
= = = =<br />
1− r ⎛ ⎛ 2 ⎞⎞ ⎛5<br />
⎞ 5<br />
⎜1−⎜− ⎟ ⎜ ⎟<br />
3<br />
⎟<br />
⎝ ⎝ ⎠⎠<br />
⎝3<br />
⎠<br />
1<br />
64. a1<br />
= 4, r =−<br />
2<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
4 4 8<br />
S∞<br />
= = = =<br />
1− r ⎛ ⎛ 1 ⎞⎞ ⎛3<br />
⎞ 3<br />
⎜1−⎜− ⎟ ⎜ ⎟<br />
2<br />
⎟<br />
⎝ ⎝ ⎠⎠<br />
⎝2<br />
⎠<br />
65.<br />
66.<br />
∞ k ∞ k−1 ∞ k−1<br />
⎛2⎞ 2 ⎛2⎞ ⎛2⎞<br />
∑3⎜ ⎟ = ∑3⋅ ⋅ ⎜ ⎟ = ∑ 2⎜ ⎟<br />
k= 1 ⎝3⎠ k= 1 3 ⎝3⎠ k=<br />
1 ⎝3⎠<br />
2<br />
a 1 = 2 , r =<br />
3<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
2 2<br />
S∞<br />
= = = = 6<br />
1−<br />
r 1−<br />
2 1<br />
3 3<br />
∞ k ∞ k−1 ∞ k−1<br />
⎛3⎞ 3 ⎛3⎞ 3⎛3⎞<br />
∑2⎜ ⎟ = ∑2⋅ ⋅ ⎜ ⎟ = ∑ ⎜ ⎟<br />
k= 1 ⎝4⎠ k= 1 4 ⎝4⎠ k=<br />
<strong>12</strong>⎝4⎠<br />
3 3<br />
a<br />
1<br />
= , r =<br />
2 4<br />
Since r < 1, <strong>the</strong> series converges.<br />
3 3<br />
a1<br />
2 2 3<br />
S∞<br />
= = = = ⋅ 4=<br />
6<br />
1−<br />
r 1−<br />
3 1 2 4 4<br />
67. { n + 2 }<br />
d = ( n+ 1+ 2) − ( n+ 2) = n+ 3−n− 2=<br />
1<br />
The difference between consecutive terms is<br />
constant. Therefore <strong>the</strong> sequence is arithmetic.<br />
50 50 50<br />
S = ( k + 2) = k+<br />
2<br />
50<br />
68. { 2n − 5}<br />
∑ ∑ ∑<br />
k= 1 k= 1 k=<br />
1<br />
50(50 + 1)<br />
= + 2(50) = <strong>12</strong>75 + 100 = 1375<br />
2<br />
d = 2( n+ 1) −5 −(2n−5)<br />
= 2n+ 2−5− 2n+ 5=<br />
2<br />
The difference between consecutive terms is<br />
constant. Therefore <strong>the</strong> sequence is arithmetic.<br />
50<br />
50 50 50<br />
∑ ∑ ∑<br />
S = (2k − 5) = 2 k−<br />
5<br />
k= 1 k= 1 k=<br />
1<br />
⎛50(50 + 1) ⎞<br />
= 2⎜<br />
⎟− 5(50) = 2550 − 250 = 2300<br />
⎝ 2 ⎠<br />
2<br />
69. { 4n } Examine <strong>the</strong> terms of <strong>the</strong> sequence: 4,<br />
16, 36, 64, 100, ...<br />
There is no common difference and <strong>the</strong>re is no<br />
common ratio. Therefore <strong>the</strong> sequence is nei<strong>the</strong>r<br />
arithmetic nor geometric.<br />
2<br />
70. { 5 1}<br />
71.<br />
72.<br />
n + Examine <strong>the</strong> terms of <strong>the</strong> sequence:<br />
6, 21, 46, 81, <strong>12</strong>6, ...<br />
There is no common difference and <strong>the</strong>re is no<br />
common ratio. Therefore <strong>the</strong> sequence is nei<strong>the</strong>r<br />
arithmetic nor geometric.<br />
⎧ 2 ⎫<br />
⎨ 3 − n ⎬<br />
⎩ 3 ⎭<br />
⎛ 2 ⎞ ⎛ 2 ⎞<br />
d = ⎜3 − ( n+ 1) ⎟−⎜3−<br />
n⎟<br />
⎝ 3 ⎠ ⎝ 3 ⎠<br />
2 2 2 2<br />
= 3− n− − 3+ n =−<br />
3 3 3 3<br />
The difference between consecutive terms is<br />
constant. Therefore <strong>the</strong> sequence is arithmetic.<br />
50 50 50<br />
⎛ 2 ⎞ 2<br />
S50<br />
= ∑⎜<br />
3− k ⎟= ∑3−<br />
∑k<br />
k= 1⎝<br />
3 ⎠ k= 1 3 k=<br />
1<br />
2 ⎛50(50 + 1) ⎞<br />
= 3(50) − ⎜ ⎟= 150 − 850 =−700<br />
3⎝<br />
2 ⎠<br />
⎧ 3 ⎫<br />
⎨ 8 − n ⎬<br />
⎩ 4 ⎭<br />
⎛ 3 ⎞ ⎛ 3 ⎞<br />
d = ⎜8 − ( n+ 1) ⎟−⎜8−<br />
n⎟<br />
⎝ 4 ⎠ ⎝ 4 ⎠<br />
3 3 3 3<br />
= 8− n− − 8+ n =−<br />
4 4 4 4<br />
The difference between consecutive terms is<br />
constant. Therefore <strong>the</strong> sequence is arithmetic.<br />
50 50 50<br />
S ⎛ 3 3<br />
50<br />
8 8<br />
1 4 k ⎞<br />
= ∑⎜<br />
− ⎟= ∑ − ∑<br />
k k 1 4<br />
k<br />
= ⎝ ⎠ = k=<br />
1<br />
3 ⎛50(50 + 1) ⎞<br />
= 8(50) − ⎜ ⎟<br />
4⎝<br />
2 ⎠<br />
= 400 − 956.25 =−556.25<br />
<strong>12</strong>58<br />
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Section <strong>12</strong>.3: Geometric <strong>Sequences</strong>; Geometric Series<br />
73. 1, 3, 6, 10, ...<br />
There is no common difference and <strong>the</strong>re is no<br />
common ratio. Therefore <strong>the</strong> sequence is nei<strong>the</strong>r<br />
arithmetic nor geometric.<br />
74. 2, 4, 6, 8, ...<br />
The common difference is 2.<br />
The difference between consecutive terms is<br />
constant. Therefore <strong>the</strong> sequence is arithmetic.<br />
50 50<br />
⎛50(50 + 1) ⎞<br />
S50<br />
= ∑( 2k ) = 2∑<br />
k = 2⎜<br />
⎟=<br />
2550<br />
k= 1 k=<br />
1 ⎝ 2 ⎠<br />
⎧<br />
n<br />
⎪⎛2<br />
⎞ ⎫⎪<br />
75. ⎨⎜<br />
⎟ ⎬<br />
⎪⎩⎝3<br />
⎠ ⎪⎭<br />
n+<br />
1<br />
⎛2<br />
⎞<br />
⎜ ⎟<br />
n+−<br />
1 n<br />
3 ⎛2⎞<br />
2<br />
r =<br />
⎝ ⎠<br />
=<br />
n ⎜ ⎟ =<br />
⎛2<br />
⎞ ⎝3⎠<br />
3<br />
⎜ ⎟<br />
⎝3<br />
⎠<br />
The ratio of consecutive terms is constant.<br />
Therefore <strong>the</strong> sequence is geometric.<br />
50<br />
⎛2<br />
⎞<br />
1<br />
50 k −<br />
2 2 ⎜ ⎟<br />
⎛ ⎞ 3<br />
S50<br />
=<br />
⎝ ⎠<br />
∑ ⎜ ⎟ = ⋅ = 1.999999997<br />
k = 1 ⎝ 3⎠ 3 2<br />
1−<br />
3<br />
⎧<br />
n<br />
⎪⎛5<br />
⎞ ⎫⎪<br />
76. ⎨⎜<br />
⎟ ⎬<br />
⎪⎩⎝4<br />
⎠ ⎪⎭<br />
n+<br />
1<br />
⎛5<br />
⎞<br />
⎜ ⎟<br />
n+−<br />
1 n<br />
4 ⎛5⎞<br />
5<br />
r =<br />
⎝ ⎠<br />
=<br />
n ⎜ ⎟ =<br />
⎛5<br />
⎞ ⎝4⎠<br />
4<br />
⎜ ⎟<br />
⎝4<br />
⎠<br />
The ratio of consecutive terms is constant.<br />
Therefore <strong>the</strong> sequence is geometric.<br />
50<br />
⎛5<br />
⎞<br />
1<br />
50 k −<br />
5 5 ⎜ ⎟<br />
⎛ ⎞ 4<br />
S50<br />
=<br />
⎝ ⎠<br />
∑ ⎜ ⎟ = ⋅ ≈350,319.6161<br />
k = 1 ⎝ 4⎠ 4 5<br />
1−<br />
4<br />
77. –1, 2, –4, 8, ...<br />
2 −4 8<br />
r = = = =−2<br />
−1 2 −4<br />
The ratio of consecutive terms is constant.<br />
Therefore <strong>the</strong> sequence is geometric.<br />
50<br />
50<br />
k −1<br />
1 −( −2)<br />
S50<br />
= ∑ −1( ⋅ − 2) = −1⋅<br />
k = 1<br />
1 − ( − 2)<br />
14<br />
≈ 3.752999689×<br />
10<br />
78. 1, 1, 2, 3, 5, 8, ...<br />
There is no common difference and <strong>the</strong>re is no<br />
common ratio. Therefore <strong>the</strong> sequence is nei<strong>the</strong>r<br />
arithmetic nor geometric.<br />
/2<br />
79. { }<br />
3 n ⎛n+<br />
1⎞<br />
⎜ ⎟<br />
2<br />
⎛n+<br />
1 n⎞<br />
3⎝ ⎠ ⎜ − ⎟<br />
⎝ 2 2⎠<br />
⎛n<br />
⎞<br />
⎜ ⎟<br />
⎝2<br />
⎠<br />
r = = 3 = 3<br />
1/2<br />
3<br />
The ratio of consecutive terms is constant.<br />
Therefore <strong>the</strong> sequence is geometric.<br />
S<br />
1−<br />
3<br />
1/2<br />
( ) 50<br />
50<br />
k /2 1/2<br />
50<br />
= ∑ 3 = 3 ⋅<br />
1/2<br />
k = 1<br />
1−<br />
3<br />
<strong>12</strong><br />
80. {( − 1) n<br />
}<br />
≈ 2.004706374×<br />
10<br />
n+<br />
1<br />
( −1)<br />
n+−<br />
1 n<br />
r = = ( − 1) = −1<br />
n<br />
( −1)<br />
The ratio of consecutive terms is constant.<br />
Therefore <strong>the</strong> sequence is geometric.<br />
S<br />
50<br />
50<br />
50<br />
k 1 −( −1)<br />
= ∑ ( − 1) = ( −1) ⋅ = 0<br />
1 −( −1)<br />
k = 1<br />
81. Find <strong>the</strong> common ratio of <strong>the</strong> terms and solve <strong>the</strong><br />
system of equations:<br />
x+ 2 x+<br />
3<br />
= r;<br />
= r<br />
x x+<br />
2<br />
x+ 2 x+<br />
3 2 2<br />
= → x + 4x+ 4= x + 3x→ x = −4<br />
x x+<br />
2<br />
82. Find <strong>the</strong> common ratio of <strong>the</strong> terms and solve <strong>the</strong><br />
system of equations:<br />
x x+<br />
2<br />
= r;<br />
= r<br />
x−1<br />
x<br />
x+ 2 x = →<br />
2 2<br />
x + x − 2 = x → x = 2<br />
x x−1<br />
83. This is a geometric series with<br />
a1 = $18,000, r = 1.05, n = 5 . Find <strong>the</strong> 5th<br />
term:<br />
−<br />
( ) ( )<br />
5 1 4<br />
a 5 = 18000 1.05 = 18000 1.05 = $21,879.11<br />
84. This is a geometric series with<br />
a1 = $15,000, r = 0.85, n = 6 . Find <strong>the</strong> 6th<br />
term:<br />
−<br />
( ) ( )<br />
6 1 5<br />
a 6 = 15000 0.85 = 15000 0.85 = $6655.58<br />
<strong>12</strong>59<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
85. a. Find <strong>the</strong> 10th term of <strong>the</strong> geometric<br />
sequence:<br />
a1<br />
= 2, r = 0.9, n=<br />
10<br />
10−1 9<br />
a = 2(0.9) = 2(0.9) = 0.775 feet<br />
10<br />
b. Find n when a n < 1:<br />
n−1<br />
2(0.9) < 1<br />
n−1<br />
( 0.9)<br />
< 0.5<br />
( ) ( )<br />
log ( 0.5)<br />
n − 1 ><br />
log ( 0.9)<br />
log ( 0.5)<br />
n<br />
log ( 0.9)<br />
( n − 1)log 0.9 < log 0.5<br />
> + 1≈7.58<br />
On <strong>the</strong> 8th swing <strong>the</strong> arc is less than 1 foot.<br />
c. Find <strong>the</strong> sum of <strong>the</strong> first 15 swings:<br />
15<br />
⎛<br />
15<br />
1 − (0.9) ⎞ ⎛1−<br />
( 0.9)<br />
⎞<br />
S15<br />
= 2 2<br />
⎜ 1 0.9 ⎟<br />
=<br />
⎝ − ⎠<br />
⎜ 0.1 ⎟<br />
⎝ ⎠<br />
15<br />
= 20 1− 0.9 = 15.88 feet<br />
( ( ) )<br />
d. Find <strong>the</strong> infinite sum of <strong>the</strong> geometric series:<br />
2 2<br />
S ∞ = = = 20 feet<br />
1−<br />
0.9 0.1<br />
86. a. Find <strong>the</strong> 3rd term of <strong>the</strong> geometric<br />
sequence:<br />
a1<br />
= 24, r = 0.8, n = 3<br />
3−1 2<br />
a = 24(0.8) = 24(0.8) = 15.36 feet<br />
3<br />
b. The height after <strong>the</strong> n th bounce is:<br />
n−1<br />
−1<br />
n<br />
a = 24(0.8) = 24 0.8 0.8<br />
n<br />
( )<br />
n<br />
= 30 0.8 ft<br />
c. Find n when a n < 0.5 :<br />
n−1<br />
24(0.8) < 0.5<br />
( ) ( )<br />
n−1<br />
( 0.8)<br />
< 0.020833<br />
( ) ( )<br />
log ( 0.020833)<br />
n − 1 ><br />
log ( 0.8)<br />
log ( 0.020833)<br />
n<br />
log ( 0.8)<br />
( n − 1)log 0.8 < log 0.020833<br />
> + 1≈18.35<br />
On <strong>the</strong> 19th bounce <strong>the</strong> height is less than<br />
0.5 feet.<br />
d. Find <strong>the</strong> infinite sum of <strong>the</strong> geometric<br />
series:<br />
24 24<br />
S ∞ = = = <strong>12</strong>0 feet on <strong>the</strong> upward<br />
1−<br />
0.8 0.2<br />
bounce.<br />
For <strong>the</strong> downward motion of <strong>the</strong> ball:<br />
30 30<br />
S ∞ = = = 150 feet<br />
1−<br />
0.8 0.2<br />
The total distance <strong>the</strong> ball travels is<br />
<strong>12</strong>0 + 150 = 270 feet.<br />
87. This is a geometric sequence with<br />
a1 = 1, r = 2, n = 64 .<br />
Find <strong>the</strong> sum of <strong>the</strong> geometric series:<br />
⎛<br />
64 64<br />
1−2 ⎞ 1−2<br />
64<br />
S64<br />
= 1 ⎜<br />
2 1<br />
1 2 ⎟<br />
= = −<br />
⎝ − ⎠ −1<br />
19<br />
= 1.845×<br />
10 grains<br />
88. This is an infinite geometric series with<br />
a 1 1<br />
1 = , r = .<br />
4 4<br />
Find <strong>the</strong> sum of <strong>the</strong> infinite geometric series:<br />
1 1<br />
( ) ( ) 1<br />
S<br />
4 4<br />
∞ = = =<br />
1 3<br />
( 1−<br />
) ( )<br />
3<br />
4 4<br />
Therefore, 1 3<br />
of <strong>the</strong> square is eventually shaded.<br />
89. The common ratio, r = 0.90 < 1. The sum is:<br />
1 1<br />
S = = = 10 .<br />
1−<br />
0.9 0.10<br />
The multiplier is 10.<br />
90. The common ratio, r = 0.95 < 1. The sum is:<br />
1 1<br />
S = = = 20 .<br />
1−<br />
0.95 0.05<br />
The multiplier is 20.<br />
91. This is an infinite geometric series with<br />
1.03<br />
a = 4, and r = .<br />
1.09<br />
Find <strong>the</strong> sum:<br />
4<br />
Price = $72.67<br />
1.03<br />
≈ per share.<br />
⎛ ⎞<br />
⎜1−<br />
⎟<br />
⎝ 1.09 ⎠<br />
<strong>12</strong>60<br />
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Section <strong>12</strong>.3: Geometric <strong>Sequences</strong>; Geometric Series<br />
92. This is an infinite geometric series with<br />
1.04<br />
a1<br />
= 2.5, and r = .<br />
1.11<br />
Find <strong>the</strong> sum:<br />
2.5<br />
Price = ≈$39.64<br />
per share.<br />
⎛ 1.04 ⎞<br />
⎜1−<br />
⎟<br />
⎝ 1.11 ⎠<br />
93. Given: a1 = 1000, r = 0.9<br />
Find n when a n < 0.01 :<br />
n−1<br />
1000(0.9) < 0.01<br />
n−1<br />
( 0.9)<br />
< 0.00001<br />
( ) ( )<br />
log ( 0.00001)<br />
n − 1 ><br />
log ( 0.9)<br />
log ( 0.00001)<br />
n<br />
log ( 0.9)<br />
( n − 1)log 0.9 < log 0.00001<br />
> + 1 ≈110.27<br />
On <strong>the</strong> 111th day or December 20, 2007, <strong>the</strong><br />
amount will be less than $0.01.<br />
Find <strong>the</strong> sum of <strong>the</strong> geometric series:<br />
111<br />
⎛<br />
n<br />
1−<br />
r ⎞ ⎛1−<br />
( 0.9)<br />
⎞<br />
S111 = a1<br />
1000⎜<br />
⎟<br />
⎜ 1 r ⎟<br />
=<br />
⎝ − ⎠<br />
⎜ 1−0.9<br />
⎟<br />
⎝ ⎠<br />
⎛<br />
111<br />
1−<br />
( 0.9)<br />
⎞<br />
= 1000 ⎜ ⎟=<br />
$9999.92<br />
⎜ 0.1 ⎟<br />
⎝ ⎠<br />
94. Both options are geometric sequences:<br />
Option A: a1 = $20,000; r = 1.06; n = 5<br />
5−1 4<br />
a5<br />
= 20,000(1.06) = 20,000(1.06) = $25,250<br />
⎛<br />
5<br />
1−<br />
( 1.06)<br />
⎞<br />
S5<br />
= 20000 ⎜ ⎟=<br />
$1<strong>12</strong>,742<br />
⎜ 1−1.06<br />
⎟<br />
⎝ ⎠<br />
Option B: a1 = $22,000; r = 1.03; n = 5<br />
5−1 4<br />
a5<br />
= 22,000(1.03) = 22,000(1.03) = $24,761<br />
⎛<br />
5<br />
1−<br />
( 1.03)<br />
⎞<br />
S5<br />
= 22000 ⎜ ⎟=<br />
$116,801<br />
⎜ 1−1.03<br />
⎟<br />
⎝ ⎠<br />
Option A provides more money in <strong>the</strong> 5th year,<br />
while Option B provides <strong>the</strong> greatest total<br />
amount of money over <strong>the</strong> 5 year period.<br />
95. Find <strong>the</strong> sum of each sequence:<br />
A: Arithmetic series with:<br />
a1 = $1000, d =− 1, n = 1000<br />
Find <strong>the</strong> sum of <strong>the</strong> arithmetic series:<br />
1000<br />
S 1000 = ( 1000 + 1 ) = 500(1001) = $500,500<br />
2<br />
B: This is a geometric sequence with<br />
a 1= 1, r = 2, n = 19.<br />
Find <strong>the</strong> sum of <strong>the</strong> geometric series:<br />
⎛<br />
19 19<br />
1−2 ⎞ 1−2<br />
19<br />
S19<br />
= 1 ⎜<br />
= = 2 − 1 = $524,287<br />
1 2 ⎟<br />
⎝ − ⎠ −1<br />
B results in more money.<br />
96. Option 1:<br />
Total Salary = $2,000,000(7) + $100,000(7)<br />
= $14,700,000<br />
Option 2: Geometric series with:<br />
a1 = $2,000,000, r = 1.045, n = 7<br />
Find <strong>the</strong> sum of <strong>the</strong> geometric series:<br />
⎛1−<br />
( 1.045) 7 ⎞<br />
S = 2,000,000 ⎜ ⎟≈$16,038,304<br />
⎜ 1−1.045<br />
⎟<br />
⎝ ⎠<br />
Option 3: Arithmetic series with:<br />
a1 = $2,000,000, d = $95,000, n = 7<br />
Find <strong>the</strong> sum of <strong>the</strong> arithmetic series:<br />
7<br />
S 7 = ( 2(2,000,000) + (7 − 1)(95,000) )<br />
2<br />
= $15,995,000<br />
Option 2 provides <strong>the</strong> most money; Option 1<br />
provides <strong>the</strong> least money.<br />
97. The amount paid each day forms a geometric<br />
sequence with a 1 = 0.01 and r = 2 .<br />
22 22<br />
1−r<br />
1−2<br />
S22 = a1<br />
⋅ = 0.01⋅ = 41,943.03<br />
1−<br />
r 1−<br />
2<br />
The total payment would be $41,943.03 if you<br />
worked all 22 days.<br />
( ) 21<br />
22 1<br />
a22 = a1 ⋅ r − = 0.01 2 = 20,971.52<br />
The payment on <strong>the</strong> 22 nd day is $20,971.52.<br />
Answers will vary. With this payment plan, <strong>the</strong><br />
bulk of <strong>the</strong> payment is at <strong>the</strong> end so missing even<br />
one day can dramatically reduce <strong>the</strong> overall<br />
payment. Notice that with one sick day you<br />
would lose <strong>the</strong> amount paid on <strong>the</strong> 22 nd day<br />
which is about half <strong>the</strong> total payment for <strong>the</strong> 22<br />
days.<br />
<strong>12</strong>61<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
98. Yes, a sequence can be both arithmetic and<br />
geometric. For example, <strong>the</strong> constant sequence<br />
3,3,3,3,..... can be viewed as an arithmetic<br />
sequence with a 1 = 3 and d = 0. Alternatively,<br />
<strong>the</strong> same sequence can be viewed as a geometric<br />
sequence with a 1 = 3 and r = 1.<br />
99 – 100. Answers will vary.<br />
101. Answers will vary. Both increase (or decrease)<br />
exponentially, but <strong>the</strong> domain of a geometric<br />
sequence is <strong>the</strong> set of natural numbers while <strong>the</strong><br />
domain of an exponential function is <strong>the</strong> set of<br />
all real numbers.<br />
Section <strong>12</strong>.4<br />
1. I: n = 1: 2⋅ 1= 2 and1(1 + 1) = 2<br />
II: If 2+ 4+ 6+ + 2 k = k( k+<br />
1) , <strong>the</strong>n<br />
2+ 4+ 6+ + 2k+ 2( k+<br />
1)<br />
= ( 2+ 4+ 6+ + 2k)<br />
+ 2( k+<br />
1)<br />
= kk ( + 1) + 2( k+<br />
1)<br />
2<br />
= k + 3k<br />
+ 2<br />
= ( k + 1)( k + 2)<br />
= ( k+ 1) ⎡⎣( k+ 1)<br />
+ 1⎤⎦<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
2. I: n = 1: 4⋅1− 3= 1and1(2⋅1− 1) = 1<br />
II: If 1+ 5+ 9 + + (4k − 3) = k(2k−1)<br />
, <strong>the</strong>n<br />
1+ 5+ 9 + + (4k− 3) + [4( k+ 1) −3]<br />
= [ 1+ 5+ 9 + + (4k− 3) ] + 4k+ 4−3<br />
= k(2k− 1) + 4k<br />
+ 1<br />
2<br />
= 2k − k+ 4k+<br />
1<br />
2<br />
= 2k<br />
+ 3k+<br />
1<br />
= ( k+ 1)(2k<br />
+ 1)<br />
= ( k+ 1) ⎡⎣2( k + 1)<br />
−1⎤⎦<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
3. I:<br />
1<br />
n = 1: 1+ 2 = 3and ⋅ 1(1 + 5) = 3<br />
2<br />
1<br />
II: If 3+ 4+ 5 + + ( k+ 2) = ⋅ k( k+<br />
5) , <strong>the</strong>n<br />
2<br />
3+ 4+ 5 + + ( k+ 2) + [( k+ 1) + 2]<br />
= [ 3+ 4+ 5 + + ( k+ 2) ] + ( k+<br />
3)<br />
1<br />
= ⋅ kk ( + 5) + ( k+<br />
3)<br />
2<br />
1 2 5<br />
= k + k + k+<br />
3<br />
2 2<br />
1 2 7<br />
= k + k + 3<br />
2 2<br />
1 2<br />
= ⋅ ( k + 7k+<br />
6)<br />
2<br />
= 1 ⋅ ( k+ 1)( k + 6)<br />
2<br />
= 1 ⋅ ( k+ 1)[( k + 1) + 5]<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
4. I: n = 1: 2⋅ 1+ 1= 3and1(1 + 2) = 3<br />
II: If 3+ 5+ 7 + + (2k+ 1) = k( k+<br />
2) , <strong>the</strong>n<br />
3 + 5 + 7 + + (2k+ 1) + [2( k+ 1) + 1]<br />
= [ 3+ 5+ 7 + + (2k+ 1) ] + (2k+<br />
3)<br />
= kk ( + 2) + (2k+<br />
3)<br />
2<br />
= k + 2k+ 2k+<br />
3<br />
2<br />
= k + 4k+<br />
3<br />
= ( k + 1)( k + 3)<br />
= ( k + 1)[( k+ 1) + 2]<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
<strong>12</strong>62<br />
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Section <strong>12</strong>.4: Ma<strong>the</strong>matical <strong>Induction</strong><br />
5. I:<br />
6. I:<br />
1<br />
n = 1: 3⋅1− 1= 2 and ⋅1(3⋅ 1+ 1) = 2<br />
2<br />
1<br />
II: If 2+ 5+ 8 + + (3k− 1) = ⋅ k(3k+<br />
1) , <strong>the</strong>n<br />
2<br />
2 + 5 + 8 + + (3k− 1) + [3( k+ 1) −1]<br />
= [ 2+ 5+ 8 + + (3k− 1) ] + (3k+<br />
2)<br />
1 3 2 1<br />
= ⋅ k(3k+ 1) + (3k+ 2) = k + k+ 3k+<br />
2<br />
2 2 2<br />
3 2 7 1 2<br />
= k + k+ 2= ( 3k + 7k+<br />
4)<br />
2 2 2<br />
= 1 ( k+ 1)(3 k+<br />
4)<br />
2<br />
1<br />
= ( k+ 1)[3( k + 1) + 1]<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
1<br />
n = 1: 3⋅1− 2 = 1and ⋅1(3⋅1− 1) = 1<br />
2<br />
1<br />
II: If 1+ 4+ 7 + + (3k− 2) = ⋅k(3k−1)<br />
,<br />
2<br />
<strong>the</strong>n<br />
1+ 4+ 7 + + (3k− 2) + [3( k+ 1) −2]<br />
= [ 1+ 4+ 7 + + (3k− 2) ] + (3k+<br />
1)<br />
1 3 2 1<br />
= ⋅k(3k − 1) + (3k+ 1) = k − k+ 3k+<br />
1<br />
2 2 2<br />
3 2 5 1 2<br />
= k + k + 1= ( 3k + 5k+<br />
2)<br />
2 2 2<br />
= 1 ( k+ 1)(3 k+<br />
2)<br />
2<br />
1<br />
= ( k+ 1)[3( k + 1) − 1]<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
7. I:<br />
11 −<br />
1<br />
n = 1: 2 = 1and 2 − 1=<br />
1<br />
2 k−1<br />
II: If 1+ 2+ 2 + + 2 = 2 −1, <strong>the</strong>n<br />
2 k−<br />
1 k+ 1−1<br />
1+ 2+ 2 + + 2 + 2<br />
2 k−1<br />
= ⎡1 2 2 2 ⎤<br />
⎣<br />
+ + + +<br />
⎦<br />
+ 2<br />
k k k<br />
= 2 − 1+ 2 = 2⋅2 −1<br />
k + 1<br />
= 2 −1<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
k<br />
k<br />
8. I:<br />
11 − 1 1<br />
n = 1: 3 = 1and (3 − 1) = 1<br />
2<br />
2 k−1 1<br />
II: If 1+ 3+ 3 + + 3 = ⋅(3 −1)<br />
, <strong>the</strong>n<br />
2<br />
2 k−<br />
1 k+ 1−1<br />
1+ 3+ 3 + + 3 + 3<br />
2 k−1<br />
k<br />
= ⎡1 3 3 3 ⎤<br />
⎣<br />
+ + + +<br />
⎦<br />
+ 3<br />
1 k k 1 k 1 k<br />
= ⋅(3 − 1) + 3 = ⋅3 − + 3<br />
2 2 2<br />
3 k 1 1 k<br />
= ⋅3 − = ⋅( 3⋅3 −1)<br />
2 2 2<br />
1 1<br />
( 3<br />
k +<br />
= − 1 )<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
11 −<br />
1<br />
9. I: n<br />
1<br />
( )<br />
= 1: 4 = 1and ⋅ 4 − 1 = 1<br />
3<br />
2 k−1 1 k<br />
II: If 1+ 4+ 4 + + 4 = ⋅( 4 −1)<br />
2 k−1<br />
( )<br />
( )<br />
3<br />
2 k−<br />
1 k+ 1−1<br />
1+ 4+ 4 + + 4 + 4<br />
= 1+ 4+ 4 + + 4 + 4<br />
1 1 1<br />
= ⋅ 4 − 1 + 4 = ⋅4 − + 4<br />
3 3 3<br />
4 k 1 1 k<br />
= ⋅4 − = ( 4⋅4 −1)<br />
3 3 3<br />
1 1<br />
( 4<br />
k +<br />
= ⋅ − 1 )<br />
3<br />
k k k k<br />
k<br />
k<br />
, <strong>the</strong>n<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
11 −<br />
1<br />
10. I: n<br />
1<br />
( )<br />
= 1: 5 = 1and ⋅ 5 − 1 = 1<br />
4<br />
2 k−1 1 k<br />
II: If 1+ 5+ 5 + + 5 = ⋅( 5 −1)<br />
4<br />
2 k−<br />
1 k+ 1−1<br />
1+ 5+ 5 + + 5 + 5<br />
2 k−1<br />
k<br />
= ⎡1 5 5 5 ⎤<br />
⎣<br />
+ + + +<br />
⎦<br />
+ 5<br />
1 1 1<br />
= ⋅( 5 − 1)<br />
+ 5 = ⋅5 − + 5<br />
4 4 4<br />
5 k 1 1 k<br />
= ⋅5 − = ( 5⋅5 −1)<br />
4 4 4<br />
1 1<br />
( 5<br />
k +<br />
= ⋅ − 1 )<br />
4<br />
k k k k<br />
, <strong>the</strong>n<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
<strong>12</strong>63<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
1 1 1 1<br />
11. I: n = 1: = and =<br />
1(1+ 1) 2 1+<br />
1 2<br />
1 1 1 1 k<br />
II: If + + + + = , <strong>the</strong>n<br />
<strong>12</strong> ⋅ 23 ⋅ 34 ⋅ kk ( + 1) k+<br />
1<br />
1 1 1 1 1<br />
+ + + + +<br />
1⋅2 2 ⋅3 3⋅ 4 kk ( + 1) ( k+ 1)( k+ 1 + 1)<br />
⎡ 1 1 1 1 ⎤ 1<br />
= ⎢ + + + + ⎥+<br />
⎣<strong>12</strong> ⋅ 23 ⋅ 34 ⋅ kk ( + 1) ⎦ ( k+ 1)( k+<br />
2)<br />
2 2<br />
k 1 k k + 2 1 k + 2k+ 1 ( k+ 1) k+ 1 k+<br />
1<br />
= + = ⋅ + = = = =<br />
k + 1 ( k + 1)( k+ 2) k+ 1 k+ 2 ( k+ 1)( k+ 2) ( k+ 1)( k+ 2) ( k+ 1)( k+ 2) k+ 2 k+ 1 + 1<br />
Conditions I and II are satisfied; <strong>the</strong> statement is true.<br />
( )<br />
<strong>12</strong>. I:<br />
1 1 1 1<br />
n = 1: = and =<br />
(2⋅1−1)(2⋅ 1+ 1) 3 2⋅ 1+<br />
1 3<br />
13. I:<br />
1 1 1 1 k<br />
II: If + + + + = , <strong>the</strong>n<br />
1⋅3 3⋅5 5⋅7 (2k− 1)(2k + 1) 2k+<br />
1<br />
1 1 1 1 1<br />
+ + + + +<br />
1⋅3 3⋅5 5 ⋅7 (2k − 1)(2k + 1) (2( k+ 1) − 1)(2( k+ 1) + 1)<br />
⎡ 1 1 1 1 ⎤ 1<br />
= ⎢ + + + + ⎥+<br />
⎣1⋅3 3⋅5 5⋅7 (2k− 1)(2k+ 1) ⎦ (2k+ 1)(2k+<br />
3)<br />
k 1 k 2k+<br />
3 1<br />
= + = ⋅ +<br />
2k + 1 (2k + 1)(2k+ 3) 2k+ 1 2k+ 3 (2k+ 1)(2k+<br />
3)<br />
2<br />
2k + 3k+ 1 ( k+ 1)(2k+ 1) k+<br />
1 k + 1<br />
= = = =<br />
(2k + 1)(2k + 3) (2k+ 1)(2k+<br />
3) 2k<br />
+ 3 2 k+ 1 + 1<br />
Conditions I and II are satisfied; <strong>the</strong> statement is true.<br />
2 1<br />
n = 1: 1 = 1and ⋅ 1(1 + 1)(2⋅ 1+ 1) = 1<br />
6<br />
2 2 2 2 1<br />
( )<br />
II: If 1 + 2 + 3 + + k = ⋅ k( k+ 1)(2k+<br />
1) , <strong>the</strong>n<br />
6<br />
2 2 2 2 2 2 2 2 2 2 1<br />
2<br />
1 + 2 + 3 + + k + ( k+ 1) = ( 1 + 2 + 3 + + k ) + ( k+ 1) = k( k+ 1)(2k+ 1) + ( k+<br />
1)<br />
6<br />
⎡1 ⎤ ⎡1 2 1 ⎤ ⎡1 2 7 ⎤ 1<br />
2<br />
= ( k + 1) ⎢ k(2k+ 1) + k+ 1 ( k 1) k k k 1 ( k 1) k k 1 ( k 1) 2k 7k<br />
6<br />
6 ⎥ = + ⎢ + + +<br />
3 6 ⎥ = + ⎢ + +<br />
3 6 ⎥ = + + +<br />
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 6<br />
1<br />
= ⋅ ( k + 1)( k+ 2)(2 k+<br />
3)<br />
6<br />
= 1 ⋅ ( k + 1)[( k+ 1) + 1][2( k+ 1) + 1]<br />
6<br />
Conditions I and II are satisfied; <strong>the</strong> statement is true.<br />
( )<br />
<strong>12</strong>64<br />
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Section <strong>12</strong>.4: Ma<strong>the</strong>matical <strong>Induction</strong><br />
3 1 2 2<br />
14. I: n = 1: 1 = 1and ⋅ 1 (1+ 1) = 1<br />
4<br />
II: If 1 3 + 2 3 + 3 3 + + k 3 = 1 k 2 ( k+<br />
1)<br />
2 , <strong>the</strong>n<br />
4<br />
1<br />
1 + 2 + 3 + + k + ( k+ 1) = ⎡1 2 3 k ⎤<br />
⎣<br />
+ + + + ⎦<br />
+ ( k+ 1) = k ( k+ 1) + ( k+<br />
1)<br />
4<br />
2 ⎡1 2 ⎤ 1 2 2<br />
= ( k + 1) k k 1 ( k 1) ⎡k 4k<br />
4⎤<br />
⎢ + +<br />
4 ⎥ = + + +<br />
⎣ ⎦ 4 ⎣ ⎦<br />
1 ( 1)<br />
2 ( 2)<br />
2<br />
= ⋅ k+ k+<br />
4<br />
1 ( 1) 2 (( 1) 1) 2<br />
= ⋅ k+ k+ +<br />
4<br />
15. I:<br />
16. I:<br />
3 3 3 3 3 3 3 3 3 3 2 2 3<br />
Conditions I and II are satisfied; <strong>the</strong> statement is true.<br />
1<br />
n = 1: 5− 1= 4 and ⋅1(9 − 1) = 4<br />
2<br />
1<br />
II: If 4+ 3+ 2 + + (5 − k) = ⋅k(9 −k)<br />
, <strong>the</strong>n<br />
2<br />
1<br />
4+ 3+ 2 + + (5 − k) + [5 − ( k+ 1)] = [ 4+ 3+ 2 + + (5 − k) ] + (4 − k) = k(9 − k) + (4 −k)<br />
2<br />
9 1 2 1 2 7 1 2<br />
= k − k + 4− k =− k + k+ 4= − ⎡k 7k<br />
8⎤<br />
2 2 2 2 2⎣<br />
− −<br />
⎦<br />
1 1 1<br />
=− ( k+ 1)( k− 8) = ( k+ 1)(8 − k) = ( k+ 1) [ 9 − ( k+<br />
1) ]<br />
2 2 2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is true.<br />
1<br />
n = 1: − (1 + 1) =−2 and − ⋅ 1(1+ 3) =− 2<br />
2<br />
1<br />
II: If −2−3−4 −<br />
− ( k+ 1) =− ⋅ k( k+<br />
3) , <strong>the</strong>n<br />
2<br />
−2−3−4 −− ( k+ 1) − (( k+ 1) + 1) = [ −2−3−4 −<br />
− ( k+ 1) ] − ( k+<br />
2)<br />
1 1 2 3 1 2 5<br />
=− ⋅ kk ( + 3) − ( k+ 2) =− k − k−k− 2=− k − k−2<br />
2 2 2 2 2<br />
1 2<br />
1<br />
=− ⋅ ⎡k 5k 4 ⎤ ( k 1)( k 4)<br />
2 ⎣<br />
+ +<br />
⎦<br />
=− ⋅ + +<br />
2<br />
1<br />
=− ⋅ ( k + 1)[( k+ 1) + 3]<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is true.<br />
<strong>12</strong>65<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
17. I:<br />
18. I:<br />
1<br />
n = 1: 1(1+ 1) = 2 and ⋅ 1(1 + 1)(1+ 2) = 2<br />
3<br />
1<br />
II: If <strong>12</strong> ⋅ + 23 ⋅ + 34 ⋅ + + kk ( + 1) = ⋅ kk ( + 1)( k+<br />
2) , <strong>the</strong>n<br />
3<br />
<strong>12</strong> ⋅ + 23 ⋅ + 34 ⋅ + + kk ( + 1) + ( k+ 1)( k+ 1+ 1) = [ <strong>12</strong> ⋅ + 23 ⋅ + 34 ⋅ + + kk ( + 1) ] + ( k+ 1)( k+<br />
2)<br />
1 ⎡1<br />
⎤<br />
= ⋅ kk ( + 1)( k+ 2) + ( k+ 1)( k+ 2) = ( k+ 1)( k+ 2) k 1<br />
3 ⎢ +<br />
3 ⎥<br />
⎣ ⎦<br />
1<br />
= ⋅ ( k + 1)( k+ 2)( k+<br />
3)<br />
3<br />
1<br />
= ⋅ ( k + 1)[( k + 1) + 1][( k+ 1) + 2]<br />
3<br />
Conditions I and II are satisfied; <strong>the</strong> statement is true.<br />
1<br />
n = 1: (2⋅1−1)(2⋅ 1) = 2and ⋅ 1(1+ 1)(4⋅1− 1) = 2<br />
3<br />
1<br />
II: If 1⋅ 2 + 3⋅ 4 + 5⋅ 6 + + (2k− 1)(2 k) = ⋅ k( k+ 1)(4k−1)<br />
, <strong>the</strong>n<br />
3<br />
1⋅ 2 + 3⋅ 4 + 5⋅ 6 + + (2k− 1)(2 k) + [2( k+ 1) − 1][2( k+<br />
1)]<br />
= [ <strong>12</strong> ⋅ + 34 ⋅ + 56 ⋅ + + (2k− 1)(2) k ] + (2k+ 1)( k+ 1)2 ⋅<br />
1 ⎡1<br />
⎤<br />
= kk ( + 1)(4k− 1) + 2( k+ 1)(2k+ 1) = ( k+ 1) k(4k 1) 2(2k<br />
1)<br />
3 ⎢ − + +<br />
3<br />
⎥<br />
⎣<br />
⎦<br />
⎡4 2 1 ⎤ 1<br />
2<br />
= ( k + 1) ⎢ k − k+ 4k+ 2 ( k 1) ( 4k k <strong>12</strong>k<br />
6)<br />
3 3 ⎥ = + − + +<br />
⎣<br />
⎦ 3<br />
1 ( 1)<br />
2<br />
1<br />
= k + ( 4 k + 11 k+<br />
6 ) = ( k+ 1)( k+ 2)(4 k+<br />
3)<br />
3<br />
3<br />
1<br />
= ( k + 1)[( k+ 1) + 1][4( k+ 1) − 1]<br />
3<br />
Conditions I and II are satisfied; <strong>the</strong> statement is true.<br />
_________________________________________________________________________________________________<br />
19. I:<br />
2<br />
n = 1: 1 + 1 = 2 is divisible by 2<br />
20. I:<br />
3<br />
n = 1: 1 + 2⋅ 1 = 3 is divisible by 3<br />
2<br />
II: If k + k is divisible by 2 , <strong>the</strong>n<br />
2 2<br />
( k + 1) + ( k + 1) = k + 2k+ 1+ k+<br />
1<br />
2<br />
= ( k + k) + (2k+<br />
2)<br />
2<br />
Since k + k is divisible by 2 and 2k + 2 is<br />
2<br />
divisible by 2, <strong>the</strong>n ( k + 1) + ( k + 1) is<br />
divisible by 2.<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
3<br />
II: If k + 2k<br />
is divisible by 3 , <strong>the</strong>n<br />
3<br />
( k+ 1) + 2( k+<br />
1)<br />
3 2<br />
= k + 3k + 3k+ 1+ 2k+<br />
2<br />
3 2<br />
= ( k + 2 k) + (3k + 3k+<br />
3)<br />
Since k<br />
2<br />
3<br />
+ 2k<br />
is divisible by 3 and<br />
3k<br />
+ 3k+ 3 is divisible by 3, <strong>the</strong>n<br />
3<br />
( k + 1) + 2( k+ 1) is divisible by 3.<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
<strong>12</strong>66<br />
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Section <strong>12</strong>.4: Ma<strong>the</strong>matical <strong>Induction</strong><br />
21. I:<br />
2<br />
n = 1: 1 − 1+ 2 = 2 is divisible by 2<br />
2<br />
II: If k − k + 2 is divisible by 2 , <strong>the</strong>n<br />
2 2<br />
( k + 1) − ( k+ 1) + 2= k + 2k+ 1−k− 1+<br />
2<br />
2<br />
= ( k − k+ 2) + (2 k)<br />
2<br />
Since k − k + 2 is divisible by 2 and 2k is<br />
2<br />
divisible by 2, <strong>the</strong>n ( k + 1) − ( k + 1) + 2 is<br />
divisible by 2.<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
22. I: n = 1: 1(1 + 1)(1 + 2) = 6 is divisible by 6<br />
23. I:<br />
24. I:<br />
II: If kk ( + 1)( k+ 2) is divisible by 6 , <strong>the</strong>n<br />
( k + 1)( k+ 1+ 1)( k+ 1+<br />
2)<br />
= ( k + 1)( k+ 2)( k+<br />
3)<br />
= kk ( + 1)( k+ 2) + 3( k+ 1)( k+<br />
2).<br />
Now, kk ( + 1)( k+<br />
2) is divisible by 6;<br />
and since ei<strong>the</strong>r k + 1 or k + 2 is even,<br />
3( k + 1)( k + 2)<br />
is divisible by 6.<br />
Thus, ( k + 1)( k+ 2)( k+<br />
3)<br />
= k k+ 1 k+ 2 + 3 k+ 1 k+<br />
2<br />
( )( ) ( )( )<br />
is divisible by 6.<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
n= 1: If x > 1<strong>the</strong>n x = x > 1.<br />
II: Assume, for some natural number k, that if<br />
k<br />
x > 1 , <strong>the</strong>n x > 1 .<br />
k + 1<br />
Then x > 1, for x > 1,<br />
k+<br />
1 k<br />
x = x ⋅ x > 1⋅ x = x > 1<br />
↑<br />
k<br />
( x > 1)<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
n= 1: If 0 < x< 1<strong>the</strong>n 0 < x < 1.<br />
II: Assume, for some natural number k, that if<br />
k<br />
0< x < 1, <strong>the</strong>n 0< x < 1.<br />
Then, for 0 < x < 1,<br />
k+<br />
1 k<br />
0< x = x ⋅ x< 1⋅ x = x<<br />
1<br />
k + 1<br />
Thus, 0 < x < 1.<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
1<br />
1<br />
25. I:<br />
1 1<br />
n= 1: a−bis a factor of a − b = a−<br />
b.<br />
II: If a− b is a factor of a − b , show that<br />
a− b is a factor of a − b .<br />
k+ 1 k+<br />
1 k k<br />
a − b = a⋅a −b⋅b<br />
= aa ⋅ −ab ⋅ + ab ⋅ −bb<br />
⋅<br />
k k k<br />
= a a − b + b ( a−b)<br />
k<br />
( )<br />
k<br />
k+ 1 k+<br />
1<br />
k k k k<br />
Since a− b is a factor of a − b and a−<br />
b<br />
is a factor of a− b, <strong>the</strong>n a− b is a factor of<br />
a<br />
k+ 1 k+<br />
1<br />
− b .<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
26. I: n = 1:<br />
21 ⋅+ 1 21 ⋅+ 1 3 3<br />
a+ bis a factor of a + b = a + b .<br />
⎡ 3 3 2 2<br />
a + b = ( a+ b)( a − ab+<br />
b )<br />
⎤<br />
⎣<br />
⎦<br />
II:<br />
k<br />
k<br />
2k+ 1 2k+<br />
1<br />
If a+ b is a factor of a + b ,<br />
show that a+ b is a factor of<br />
2( k+ 1) + 1 2( k+ 1)<br />
+ 1<br />
a + b .<br />
2( k+ 1) + 1 2( k+ 1) + 1 2k+ 3 2k+<br />
3<br />
a + b = a + b<br />
= a ⋅ a + a ⋅b −a ⋅ b + b ⋅b<br />
2 2k+ 1 2k+ 1 2k+<br />
1 2 2<br />
= a a + b −b ( a −b<br />
)<br />
2 2k+ 1 2 2k+ 1 2 2k+ 1 2 2k+<br />
1<br />
( )<br />
2k+ 1 2k+<br />
1<br />
Since a+ b is a factor of a + b and<br />
2 2<br />
a+ b is a factor of a − b [( a− b)( a+ b)<br />
] ,<br />
2k+ 3 2k+<br />
3<br />
<strong>the</strong>n a+ b is a factor of a + b .<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
27. I: 1<br />
n = : ( ) 1<br />
1+ a = 1+ a ≥ 1+ 1⋅<br />
a<br />
II: Assume that <strong>the</strong>re is an integer k for which<br />
<strong>the</strong> inequality holds. We need to show that if<br />
k<br />
1+ a ≥ 1+ ka <strong>the</strong>n<br />
( )<br />
k + 1<br />
( 1+ a) ≥ 1+ ( k+ 1)<br />
a.<br />
k+<br />
1<br />
k<br />
( 1+ a) = ( 1+ a) ( 1+<br />
a)<br />
≥ ( 1+ ka)( 1+<br />
a)<br />
2<br />
= 1+ ka + a + ka<br />
= 1+ k+ 1 a+<br />
ka<br />
( )<br />
( k )<br />
≥ 1+ + 1 a<br />
Conditions I and II are satisfied, <strong>the</strong> statement is<br />
true.<br />
2<br />
<strong>12</strong>67<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
28.<br />
n = 1:<br />
2<br />
1 − 1+ 41 = 41 is a prime number.<br />
n = 41:<br />
2 2<br />
41 − 41+ 41 = 41 is not a prime number.<br />
29. II: If 2+ 4+ 6+ + 2k = k + k + 2, <strong>the</strong>n<br />
2+ 4+ 6+ + 2k+ 2( k+<br />
1)<br />
= 2+ 4+ 6+ + 2k<br />
+ 2k+<br />
2<br />
I:<br />
30. I:<br />
[ ]<br />
2<br />
= k + k+ 2+ 2k+<br />
2<br />
2<br />
= ( k + 2k+ 1) + ( k+ 1) + 2<br />
2<br />
= ( k + 1) + ( k+ 1) + 2<br />
2<br />
n = 1: 2⋅ 1= 2 and1 + 1+ 2 = 4 ≠ 2<br />
1<br />
11 −<br />
⎛1−<br />
r ⎞<br />
n= 1: ar = a and a ⎜<br />
= a<br />
1−<br />
r ⎟<br />
⎝ ⎠<br />
k<br />
2 k −1 ⎛1−<br />
r ⎞<br />
II: If a+ ar+ ar + + ar = a<br />
⎜<br />
1−<br />
r ⎟<br />
,<br />
⎝ ⎠<br />
<strong>the</strong>n<br />
2 k−<br />
1 k+ 1−1<br />
a+ ar+ ar + + ar + ar<br />
2 k−1<br />
k<br />
= ⎡a ar ar ar ⎤<br />
⎣<br />
+ + + +<br />
⎦<br />
+ ar<br />
⎛<br />
k k k<br />
1 −r ⎞<br />
k a(1 − r ) + ar (1 −r)<br />
= a<br />
⎜<br />
ar<br />
1 r ⎟<br />
+ =<br />
⎝ − ⎠<br />
1−r<br />
k k k+ 1 k+<br />
1<br />
a− ar + ar −ar ⎛1−r<br />
⎞<br />
= = a<br />
1−r<br />
⎜ 1−r<br />
⎟<br />
⎝ ⎠<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
31. I: n = 1:<br />
1(1 −1)<br />
a+ (1− 1) d = a and 1⋅ a+ d = a<br />
2<br />
II: If a+ ( a+ d) + ( a+ 2 d) + + [ a+ ( k−1)<br />
d]<br />
kk ( −1)<br />
= ka + d<br />
2<br />
<strong>the</strong>n<br />
a+ ( a+ d) + ( a+ 2 d) + + [ a+ ( k − 1) d] + ( a+<br />
kd)<br />
= [ a+ ( a+ d) + ( a+ 2 d) + + [ a+ ( k − 1) d]<br />
] + ( a+<br />
kd)<br />
kk ( −1)<br />
= ka + d + ( a + kd)<br />
2<br />
⎡kk<br />
( −1)<br />
⎤<br />
= ( k+ 1) a+ d ⎢ + k<br />
2 ⎥<br />
⎣ ⎦<br />
2<br />
⎡<br />
2<br />
k − k + 2k⎤<br />
= ( k+ 1) a+<br />
d<br />
⎢⎣<br />
2 ⎥⎦<br />
⎡<br />
2<br />
k + k⎤<br />
= ( k+ 1) a+<br />
d<br />
⎢⎣<br />
2 ⎥⎦<br />
⎡( k + 1) k⎤<br />
= ( k+ 1) a+<br />
d⎢<br />
2<br />
⎥<br />
⎣ ⎦<br />
⎡( k + 1) ⎡ ( k + 1)<br />
−1⎤⎤<br />
= ( k + 1)<br />
a+<br />
d<br />
⎣ ⎦<br />
⎢<br />
⎥<br />
⎢⎣<br />
2 ⎥⎦<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
32. I: n = 4: The number of diagonals of a<br />
quadrilateral is 1 4(4 3) 2<br />
2 ⋅ − = .<br />
II: Assume that for any integer k, <strong>the</strong> number of<br />
diagonals of a convex polygon with k sides<br />
(k vertices) is 1 ( 3)<br />
2 ⋅ kk−<br />
. A convex<br />
polygon with k + 1 sides ( k + 1 vertices)<br />
consists of a convex polygon with k sides<br />
(k vertices) plus a triangle, for a total of<br />
( k + 1) vertices. The diagonals of this<br />
k + 1-sided convex polygon consist of <strong>the</strong><br />
diagonals of <strong>the</strong> k-sided polygon plus k − 1<br />
additional diagonals. For example, consider<br />
<strong>the</strong> following diagrams.<br />
k + 1 = 6 sides<br />
k = 5 sides k − 1 = 4 new diagonals<br />
Thus, we have <strong>the</strong> equation:<br />
1 1 2 3<br />
kk ( − 3) + ( k− 1) = k − k+ k−1<br />
2 2 2<br />
1 2 1<br />
= k − k−1<br />
2 2<br />
1 2<br />
= ( k −k−2)<br />
2<br />
1<br />
= ( k + 1)( k−<br />
2)<br />
2<br />
1<br />
= ( k + 1)[( k+ 1) − 3]<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong><br />
statement is true.<br />
<strong>12</strong>68<br />
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Section <strong>12</strong>.5: The <strong>Binomial</strong> <strong>Theorem</strong><br />
33. I: n = 3: (3 −2) ⋅ 180°= 180 ° which is <strong>the</strong><br />
sum of <strong>the</strong> angles of a triangle.<br />
II: Assume that for any integer k, <strong>the</strong> sum of<br />
<strong>the</strong> angles of a convex polygon with k sides<br />
is ( k −2) ⋅ 180°. A convex polygon with<br />
k + 1 sides consists of a convex polygon<br />
with k sides plus a triangle. Thus, <strong>the</strong> sum of<br />
<strong>the</strong> angles is<br />
( k −2) ⋅ 180°+ 180 °= [( k+ 1) −2] ⋅ 180 ° .<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
34. Answers will vary.<br />
Section <strong>12</strong>.5<br />
1. Pascal Triangle<br />
2.<br />
⎛6⎞ 6! 654321 ⋅ ⋅ ⋅ ⋅ ⋅ 65 ⋅<br />
⎜ ⎟= = = = 15<br />
⎝2⎠<br />
2!4! 2⋅1⋅4⋅3⋅2⋅1 2⋅1<br />
⎛n⎞ n!<br />
3. False; ⎜ ⎟ =<br />
⎝ j ⎠ j! ( n−<br />
j)<br />
!<br />
4. <strong>Binomial</strong> <strong>Theorem</strong><br />
5.<br />
6.<br />
⎛5⎞ 5! 54321 ⋅ ⋅ ⋅ ⋅ 54 ⋅<br />
⎜ ⎟= = = = 10<br />
⎝3⎠<br />
3! 2! 3⋅2⋅1⋅2⋅1 2⋅1<br />
⎛7⎞ 7! 7654321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 765 ⋅ ⋅<br />
⎜ ⎟= = = = 35<br />
⎝3⎠<br />
3!4! 3214321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 321 ⋅ ⋅<br />
7.<br />
8.<br />
9.<br />
10.<br />
11.<br />
<strong>12</strong>.<br />
13.<br />
14.<br />
15.<br />
16.<br />
⎛7⎞ 7! 7⋅6⋅5⋅4⋅3⋅2⋅1 7⋅6<br />
⎜ ⎟= = = = 21<br />
⎝5⎠<br />
5!2! 5432<strong>12</strong>1 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 21 ⋅<br />
⎛9⎞ 9! 9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 9⋅8<br />
⎜ ⎟= = = = 36<br />
⎝7⎠<br />
7!2! 765432<strong>12</strong>1 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 21 ⋅<br />
⎛50⎞ 50! 50⋅<br />
49! 50<br />
⎜ ⎟= = = = 50<br />
⎝49⎠<br />
49!1! 49! ⋅1 1<br />
⎛100⎞ 100! 100⋅99⋅98! 100⋅99<br />
⎜ ⎟= = = = 4950<br />
⎝ 98 ⎠ 98!2! 98!21 ⋅ ⋅ 21 ⋅<br />
⎛1000⎞ 1000! 1<br />
⎜ ⎟ = = = 1<br />
⎝1000⎠<br />
1000!0! 1<br />
⎛1000⎞ 1000! 1<br />
⎜ ⎟ = = = 1<br />
⎝ 0 ⎠ 0!1000! 1<br />
⎛55⎞ 55!<br />
⎜ ⎟ = ≈ 1.8664 × 10<br />
⎝23⎠<br />
23!32!<br />
⎛60⎞ 60!<br />
⎜ ⎟ = ≈ 4.1918 × 10<br />
⎝20⎠<br />
20! 40!<br />
⎛47⎞ 47!<br />
⎜ ⎟ = ≈ 1.4834 × 10<br />
⎝25⎠<br />
25! 22!<br />
⎛37⎞ 37!<br />
⎜ ⎟ = ≈ 1.7673 × 10<br />
⎝19 ⎠ 19!18!<br />
15<br />
15<br />
13<br />
10<br />
_________________________________________________________________________________________________<br />
17.<br />
18.<br />
19.<br />
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞<br />
( x + 1) = ⎜ ⎟x + ⎜ ⎟x + ⎜ ⎟x + ⎜ ⎟x + ⎜ ⎟x + ⎜ ⎟x<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠<br />
5 5 4 3 2 1 0<br />
5 4 3 2<br />
= x + 5x + 10x + 10x + 5x+<br />
1<br />
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞<br />
( x − 1) = ⎜ ⎟x + ⎜ ⎟( − 1) x + ⎜ ⎟( − 1) x + ⎜ ⎟( − 1) x + ⎜ ⎟( − 1) x + ⎜ ⎟( −1)<br />
x<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠<br />
5 5 4 2 3 3 2 4 1 5 0<br />
5 4 3 2<br />
= x − 5x + 10x − 10x + 5x−1<br />
⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞<br />
( x− 2) = ⎜ ⎟x + ⎜ ⎟x ( − 2) + ⎜ ⎟x ( − 2) + ⎜ ⎟x ( − 2) + ⎜ ⎟x ( − 2) + ⎜ ⎟x( − 2) + ⎜ ⎟x<br />
( −2)<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ ⎝6⎠<br />
6 6 5 4 2 3 3 2 4 5 0 6<br />
6 5 4 3 2<br />
= x + 6 x ( − 2) + 15x ⋅ 4 + 20 x ( − 8) + 15x ⋅ 16 + 6 x⋅( − 32) + 64<br />
6 5 4 3 2<br />
= x − <strong>12</strong>x + 60x − 160x + 240x − 192x+<br />
64<br />
<strong>12</strong>69<br />
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as <strong>the</strong>y currently<br />
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
20.<br />
21.<br />
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞<br />
( x+ 3) = ⎜ ⎟x + ⎜ ⎟x (3) + ⎜ ⎟x (3) + ⎜ ⎟x (3) + ⎜ ⎟x (3) + ⎜ ⎟x<br />
(3)<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠<br />
5 5 4 3 2 2 3 1 4 0 5<br />
5 4 3 2<br />
= x + 5 x (3) + 10x ⋅ 9 + 10 x (27) + 5x⋅ 81+<br />
243<br />
5 4 3 2<br />
= x + 15x + 90x + 270x + 405x+<br />
243<br />
4 ⎛4⎞ 4 ⎛4⎞ 3 ⎛4⎞ 2 ⎛4⎞ ⎛4⎞<br />
(3x+ 1) = ⎜ ⎟(3 x) + ⎜ ⎟(3 x) + ⎜ ⎟(3 x) + ⎜ ⎟(3 x)<br />
+ ⎜ ⎟<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠<br />
4 3 2 4 3 2<br />
= 81x + 4⋅ 27x + 6⋅ 9x + 4⋅ 3x+ 1 = 81x + 108x + 54x + <strong>12</strong>x+<br />
1<br />
22.<br />
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞<br />
(2x+ 3) = ⎜ ⎟(2 x) + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟⋅2x⋅ 3 + ⎜ ⎟⋅3<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠<br />
5 5 4 3 2 2 3 4 5<br />
5 4 3 2<br />
= 32x + 5⋅16x ⋅ 3 + 10⋅8x ⋅ 9 + 10⋅4x ⋅ 27 + 5⋅2x⋅ 81+<br />
243<br />
5 4 3 2<br />
= 32x + 240x + 720x + 1080x + 810x+<br />
243<br />
2 2 ⎛5⎞ 2 ⎛5⎞ 2 2 ⎛5⎞ 2 2 ⎛5⎞ 2 2 ⎛5⎞ 2 2 ⎛5⎞<br />
2<br />
23. ( x + y ) = ⎜ ⎟( x ) + ⎜ ⎟( x ) y + ⎜ ⎟( x ) ( y ) + ⎜ ⎟( x ) ( y ) + ⎜ ⎟x ( y ) + ⎜ ⎟( y )<br />
5 5 4 3 2 2 3 4 5<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠<br />
10 8 2 6 4 4 6 2 8 10<br />
= x + 5x y + 10x y + 10x y + 5x y + y<br />
2 2 ⎛6⎞ 2 ⎛6⎞ 2 2 ⎛6⎞ 2 2 ⎛6⎞ 2 2 ⎛6⎞<br />
2 2<br />
24. ( x − y ) = ⎜ ⎟( x ) + ⎜ ⎟( x ) ( − y ) + ⎜ ⎟( x ) ( − y ) + ⎜ ⎟( x ) ( − y ) + ⎜ ⎟( x ) ( −y<br />
)<br />
6 6 5 4 2 3 3 2 4<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠<br />
⎛6⎞<br />
2 2<br />
5 ⎛6⎞<br />
2<br />
6<br />
+ ⎜ ⎟x ( − y ) + ⎜ ⎟( −y<br />
)<br />
⎝5⎠<br />
⎝6⎠<br />
<strong>12</strong> 10 2 8 4 6 6 4 8 2 10 <strong>12</strong><br />
= x − 6x y + 15x y − 20x y + 15x y − 6x y + y<br />
⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞<br />
25. ( x + 2) = ⎜ ⎟( x) + ⎜ ⎟( x) ( 2) + ⎜ ⎟( x) ( 2) + ⎜ ⎟( x) ( 2)<br />
6 6 5 1 4 2 3 3<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠<br />
⎛6⎞ 2 4 ⎛6⎞ 5 ⎛6⎞<br />
6<br />
+ ⎜ ⎟( x) ( 2) + ⎜ ⎟( x)( 2) + ⎜ ⎟( 2)<br />
⎝4⎠ ⎝5⎠ ⎝6⎠<br />
3 5/2 2 3/2 1/2<br />
= x + 6 2x + 15⋅ 2x + 20⋅ 2 2x + 15⋅ 4x+ 6⋅ 4 2x<br />
+ 8<br />
3 5/2 2 3/2 1/2<br />
= x + 6 2x + 30x + 40 2x + 60x+ 24 2x<br />
+ 8<br />
⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞<br />
26. ( x − 3) = ⎜ ⎟( x) + ⎜ ⎟( x) ( − 3) + ⎜ ⎟( x) ( − 3) + ⎜ ⎟( x)( − 3) + ⎜ ⎟( − 3)<br />
4 4 3 1 2 2 3 4<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠<br />
2 3/2 1/2<br />
= x − 4 3x + 63 ⋅ x−43 ⋅ 3x<br />
+ 9<br />
2 3/2 1/2<br />
= x − 4 3x + 18x− <strong>12</strong> 3x<br />
+ 9<br />
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞<br />
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠<br />
5 5 4 3 2 2 3 4 5<br />
27. ( ax + by) = ( ax) + ( ax) ⋅ by + ( ax) ( by) + ( ax) ( by) + ax( by) + ( by)<br />
5 5 4 4 3 3 2 2 2 2 3 3 4 4 5 5<br />
= a x + 5a x by+ 10a x b y + 10a x b y + 5axb y + b y<br />
⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞<br />
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠<br />
4 4 3 2 2 3 4<br />
28. ( ax − by) = ( ax) + ( ax) ( − by)<br />
+ ( ax) ( − by) + ( ax)( − by) + ( −by)<br />
4 4 3 3 2 2 2 2 3 3 4 4<br />
= a x − 4a x by + 6a x b y − 4axb y + b y<br />
<strong>12</strong>70<br />
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as <strong>the</strong>y currently<br />
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
Section <strong>12</strong>.5: The <strong>Binomial</strong> <strong>Theorem</strong><br />
29. n = 10, j = 4, x = x, a = 3<br />
⎛10⎞ 10! 10987 ⋅ ⋅ ⋅<br />
⎜ ⎟x ⋅ 3 = ⋅ 81x = ⋅81x<br />
⎝ 4 ⎠ 4!6! 4⋅3⋅2⋅1<br />
6<br />
= 17,010x<br />
6<br />
The coefficient of x is 17,010.<br />
6 4 6 6<br />
30. n= 10, j = 7, x = x, a = − 3<br />
⎛10⎞<br />
3 7 10!<br />
3<br />
⎜ ⎟x<br />
⋅− ( 3) = ⋅− ( 2187)<br />
x<br />
⎝ 7 ⎠ 7!3!<br />
10⋅9⋅8<br />
= ⋅( −2187)<br />
x<br />
321 ⋅ ⋅<br />
3<br />
=−262,440x<br />
3<br />
The coefficient of x is − 262,440.<br />
31. n = <strong>12</strong>, j = 5, x = 2 x, a = − 1<br />
⎛<strong>12</strong>⎞<br />
7 5 <strong>12</strong>! 7<br />
⎜ ⎟(2 x) ⋅− ( 1) = ⋅<strong>12</strong>8 x ( −1)<br />
⎝ 5 ⎠<br />
5!7!<br />
<strong>12</strong>⋅11⋅10⋅9⋅8 = ⋅ ( − <strong>12</strong>8) x<br />
54321 ⋅ ⋅ ⋅ ⋅<br />
7<br />
=−101,376x<br />
7<br />
The coefficient of x is − 101,376.<br />
32. n= <strong>12</strong>, j = 9, x = 2 x, a = 1<br />
⎛<strong>12</strong>⎞ <strong>12</strong>! <strong>12</strong>⋅11⋅10<br />
⎜ ⎟(2 x) ⋅ (1) = ⋅ 8 x (1) = ⋅8x<br />
⎝ 9 ⎠<br />
9!3! 3⋅2⋅1<br />
3<br />
= 1760x<br />
3<br />
The coefficient of x is 1760.<br />
3 9 3 3<br />
33. n= 9, j = 2, x = 2 x, a = 3<br />
⎛9⎞<br />
7 2 9! 7<br />
⎜ ⎟(2 x) ⋅ 3 = ⋅<strong>12</strong>8 x (9)<br />
⎝2⎠<br />
2!7!<br />
98 ⋅ <strong>12</strong>8<br />
7<br />
= ⋅ x ⋅ 9<br />
21 ⋅<br />
7<br />
= 41,472x<br />
7<br />
The coefficient of x is 41,472.<br />
34. n= 9, j = 7, x = 2 x, a = − 3<br />
⎛9⎞<br />
2 7 9! 2<br />
⎜ ⎟(2 x) ⋅− ( 3) = ⋅4 x ( −2187)<br />
⎝7⎠<br />
7!2!<br />
98 ⋅ 4<br />
2<br />
= ⋅ x ⋅− 2187<br />
21 ⋅<br />
2<br />
=−314,928x<br />
2<br />
The coefficient of x is − 314,928.<br />
3<br />
7<br />
35. n= 7, j = 4, x = x, a = 3<br />
⎛7⎞<br />
7! 7⋅6⋅5<br />
⎜ ⎟x ⋅ 3 = ⋅ 81x = ⋅ 81x = 2835x<br />
⎝4⎠<br />
4!3! 3⋅2⋅1<br />
3 4 3 3 3<br />
36. n= 7, j = 2, x = x, a = − 3<br />
⎛7⎞<br />
7! 7⋅6<br />
⎜ ⎟x ⋅− ( 3) = ⋅ 9x = ⋅ 9x = 189x<br />
⎝2⎠<br />
2!5! 2⋅1<br />
5 2 5 5 5<br />
37. n= 9, j = 2, x = 3 x, a =− 2<br />
⎛9⎞<br />
7 2 9!<br />
7<br />
⎜ ⎟(3 x) ⋅− ( 2) = ⋅2187x<br />
⋅4<br />
⎝2⎠<br />
2!7!<br />
98 ⋅<br />
= ⋅ 8748 x = 314,928 x<br />
21 ⋅<br />
38. n= 8, j = 5, x = 3 x, a = 2<br />
7 7<br />
⎛8⎞<br />
3 5 8! 3<br />
⎜ ⎟(3 x) ⋅ ( 2) = ⋅27x<br />
⋅32<br />
⎝5⎠<br />
5!3!<br />
876 ⋅ ⋅<br />
= ⋅ 864 x = 48,384 x<br />
321 ⋅ ⋅<br />
0<br />
39. The x term in<br />
∑<br />
<strong>12</strong> <strong>12</strong><br />
⎛<strong>12</strong>⎞ <strong>12</strong>−<br />
j ⎛1 ⎞ <strong>12</strong><br />
( )<br />
j ⎛ ⎞<br />
⎜ ⎟ x<br />
x<br />
j= 0<br />
j<br />
⎜ ⎟ = ∑ ⎜ ⎟<br />
⎝ ⎠ ⎝ x ⎠ j=<br />
0⎝ j ⎠<br />
occurs when:<br />
24 − 3 j = 0<br />
24 = 3 j<br />
j = 8<br />
3 3<br />
2 24−3j<br />
⎛<strong>12</strong>⎞ <strong>12</strong>! <strong>12</strong>⋅11⋅10 ⋅9<br />
The coefficient is ⎜ 495<br />
8 ⎟ = = =<br />
⎝ ⎠ 8!4! 4321 ⋅ ⋅ ⋅<br />
0<br />
40. The x term in<br />
∑<br />
9 j 9<br />
⎛9⎞ − ⎛ 1 ⎞ ⎛9⎞<br />
⎜ ⎟ x<br />
1<br />
2<br />
j= 0<br />
j<br />
⎜− ⎟ = ∑ ⎜ ⎟ −<br />
⎝ ⎠ ⎝ x ⎠ j=<br />
0⎝ j⎠<br />
occurs when:<br />
9− 3j<br />
= 0<br />
9=<br />
3j<br />
j = 3<br />
The coefficient is<br />
⎛9⎞<br />
( ) 3 9! 9⋅8⋅7<br />
⎜ ⎟ − 1 =− =− =−84<br />
⎝3⎠<br />
3!6! 3⋅2⋅1<br />
( ) ( )<br />
9 j j 9−3j<br />
x<br />
<strong>12</strong>71<br />
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as <strong>the</strong>y currently<br />
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
4<br />
2<br />
41. The x term in<br />
42. The x term in<br />
10 j<br />
∑ 10 3<br />
⎛10⎞ 10−<br />
j ⎛ − 2 ⎞ ⎛10⎞<br />
j 10−<br />
j<br />
( x) ( 2)<br />
x 2<br />
⎜ ⎟ ⎜ ⎟ = ∑<br />
8 j<br />
⎜ ⎟ −<br />
∑ ⎛8⎞ 8−<br />
j ⎛ 3 ⎞<br />
( ) 8 ⎛8⎞<br />
j 4−<br />
j<br />
⎜ ⎟ x ⎜ ⎟ = ∑ ⎜ ⎟( 3)<br />
x<br />
j= 0⎝ j ⎠ ⎝ x ⎠ j=<br />
0⎝ j ⎠<br />
j= 0⎝ j⎠ ⎝ x ⎠ j=<br />
0⎝ j⎠<br />
occurs when:<br />
occurs when:<br />
3<br />
4− j = 2<br />
10 − j = 4<br />
2<br />
− j =−2<br />
3<br />
− j =−6<br />
j = 2<br />
2<br />
The coefficient is<br />
j = 4<br />
⎛8⎞<br />
The coefficient is<br />
( ) 2 8! 8⋅7<br />
⎜ ⎟ 3 = ⋅ 9= ⋅ 9=<br />
252<br />
⎝2⎠<br />
6!2! 2⋅1<br />
⎛10⎞ ( ) 4 10! 10987 ⋅ ⋅ ⋅<br />
⎜ ⎟ − 2 = ⋅ 16 = ⋅ 16 = 3360<br />
⎝ 4 ⎠ 6!4! 4⋅3⋅2⋅1<br />
_________________________________________________________________________________________________<br />
5 −3 ⎛5⎞ 5 ⎛5⎞ 4 −3 ⎛5⎞ 3 −3 ⎛5⎞<br />
2 −3<br />
43. (1.001) ( 1 10 ) ⎜ ⎟1 ⎜ ⎟1 10 ⎜ ⎟1 ( 10 ) ⎜ ⎟1 ( 10 )<br />
5 2 3<br />
= + = + ⋅ + ⋅ + ⋅ +⋅⋅⋅<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠<br />
= 1+ 5(0.001) + 10(0.000001) + 10(0.000000001) +⋅⋅⋅<br />
= 1+ 0.005 + 0.000010 + 0.000000010 +⋅⋅⋅<br />
= 1.00501 (correct to 5 decimal places)<br />
6 ⎛6⎞ ⎛6⎞ ⎛6⎞ 2 ⎛6⎞<br />
3<br />
= − = ⎜ ⎟ + ⎜ ⎟ ⋅ − + ⎜ ⎟ ⋅ − + ⎜ ⎟⋅ ⋅ − +⋅⋅⋅<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠<br />
= 1+ 6( − 0.002) + 15( 0.000004) + 20( − 0.000000008 ) + ...<br />
= 1− 0.0<strong>12</strong> + 0.00006 −0.00000016<br />
= 0.98806 (correct to 5 decimal places)<br />
6 6 5 4 3<br />
44. (0.998) ( 1 0.002) 1 1 ( 0.002) 1 ( 0.002) 1 ( 0.002)<br />
45.<br />
( − )<br />
( )<br />
⎛ n ⎞ n! n!<br />
n n 1!<br />
⎜ ⎟ = = = = n ;<br />
⎝n −1⎠<br />
( n−1!( ) n−( n−1)! ) ( n−1!(1)! ) n−1!<br />
⎛n⎞ n! n! n! n!<br />
⎜ ⎟ = = = = = 1<br />
⎝n⎠<br />
n!( n−n)! n! 0! n! ⋅1 n!<br />
46.<br />
⎛ n ⎞ n! n! n!<br />
⎛n⎞<br />
⎜ ⎟= = = = ⎜ ⎟<br />
⎝n− j⎠ ( n− j)!( n−( n− j))! ( n− j)<br />
! j! j!( n−<br />
j)!<br />
⎝ j⎠<br />
⎛n⎞ ⎛n⎞ ⎛n⎞<br />
n<br />
47. Show that ⎜ ⎟+ ⎜ ⎟+⋅⋅⋅+ ⎜ ⎟=<br />
2<br />
⎝0⎠ ⎝1⎠ ⎝n⎠<br />
n n ⎛n⎞ n ⎛n⎞ n−1 ⎛n⎞ n−2 2 ⎛n⎞ n−n n ⎛n⎞ ⎛n⎞ ⎛n⎞<br />
2 = (1+ 1) = ⎜ ⎟⋅ 1 + ⎜ ⎟⋅1 ⋅ 1+ ⎜ ⎟⋅1 ⋅ 1 +⋅⋅⋅+ ⎜ ⎟⋅1 ⋅ 1 = ⎜ ⎟+ ⎜ ⎟+⋅⋅⋅+<br />
⎜ ⎟<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝n⎠ ⎝0⎠ ⎝1⎠ ⎝n⎠<br />
⎛n⎞ ⎛n⎞ ⎛n⎞ n ⎛n⎞<br />
48. Show that ⎜ ⎟− ⎜ ⎟+ ⎜ ⎟−⋅⋅⋅+ ( − 1) ⎜ ⎟=<br />
0<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝n⎠<br />
n ⎛n⎞ n ⎛n⎞ n−1 ⎛n⎞ n−2 2 ⎛n⎞ n−n n ⎛n⎞ ⎛n⎞ ⎛n⎞ n⎛n⎞<br />
0 = (1 − 1) = ⎜ ⎟⋅ 1 + ⎜ ⎟⋅1 ⋅− ( 1) + ⎜ ⎟⋅1 ⋅− ( 1) +⋅⋅⋅+ ⎜ ⎟⋅1 ⋅− ( 1) = ⎜ ⎟− ⎜ ⎟+ ⎜ ⎟−⋅⋅⋅+−<br />
( 1) ⎜ ⎟<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝n⎠ ⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝n⎠<br />
49.<br />
5 4 3 2 2 3 4 5 5<br />
⎛5⎞⎛1⎞ ⎛5⎞⎛1⎞ ⎛3⎞ ⎛5⎞⎛1⎞ ⎛3⎞ ⎛5⎞⎛1⎞ ⎛3⎞ ⎛5⎞⎛1⎞⎛3⎞ ⎛5⎞⎛3⎞ ⎛1 3⎞<br />
5<br />
⎜ ⎟ (1) 1<br />
0<br />
⎜ ⎟ + ⎜ ⎟<br />
4 1<br />
⎜ ⎟ ⎜ ⎟+ ⎜ ⎟ + + + = + = =<br />
4 4 2<br />
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟<br />
4 4 3<br />
⎜ ⎟ ⎜ ⎟<br />
4 4 4<br />
⎜ ⎟⎜ ⎟<br />
4 4 5<br />
⎜ ⎟ ⎜ ⎟<br />
⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝4⎠<br />
⎝4 4⎠<br />
<strong>12</strong>72<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
<strong>Chapter</strong> <strong>12</strong> Review Exercises<br />
50.<br />
8<br />
<strong>12</strong>! = 479,001,600 = 4.790016×<br />
10<br />
18<br />
20! ≈ 2.432902008×<br />
10<br />
25<br />
25! ≈ 1.551<strong>12</strong>1004×<br />
10<br />
⎛<strong>12</strong> ⎞ ⎛ 1 ⎞<br />
<strong>12</strong>! ≈ 2⋅<strong>12</strong>π ⎜ ⎟ ⎜1+<br />
⎟<br />
⎝ e ⎠ ⎝ <strong>12</strong> ⋅ <strong>12</strong> − 1 ⎠<br />
≈ 479,013,972.4<br />
⎛20 ⎞ ⎛ 1 ⎞<br />
20! ≈ 2⋅20π ⎜ ⎟ ⎜1+<br />
⎟<br />
⎝ e ⎠ ⎝ <strong>12</strong> ⋅ 20 − 1 ⎠<br />
18<br />
≈ 2.43292403×<br />
10<br />
⎛25 ⎞ ⎛ 1 ⎞<br />
25! ≈ 2⋅25π ⎜ ⎟ ⎜1+<br />
⎟<br />
⎝ e ⎠ ⎝ <strong>12</strong> ⋅ 25 − 1 ⎠<br />
25<br />
≈ 1.551<strong>12</strong>9917×<br />
10<br />
<strong>Chapter</strong> <strong>12</strong> Review Exercises<br />
1.<br />
2.<br />
3.<br />
4.<br />
<strong>12</strong><br />
20<br />
25<br />
11+ 3 4 2 2+<br />
3 5<br />
a1 = ( − 1) =− , a2<br />
= ( − 1) = ,<br />
1+ 2 3 2+<br />
2 4<br />
3 3+ 3 6 4 4+<br />
3 7<br />
a3 = ( − 1) =− , a4<br />
= ( − 1) = ,<br />
3+ 2 5 4+<br />
2 6<br />
5 5+<br />
3 8<br />
a5<br />
= ( − 1) =− 5 + 2 7<br />
11 +<br />
b1<br />
= ( −1) (2 ⋅ 1+ 3) = 5,<br />
2+<br />
1<br />
b2<br />
= ( −1) (2⋅ 2+ 3) =−7,<br />
3+<br />
1<br />
b3<br />
= ( −1) (2⋅ 3+ 3) = 9,<br />
4+<br />
1<br />
b4<br />
= ( −1) (2⋅ 4+ 3) =−11,<br />
5+<br />
1<br />
b = ( −1) (2⋅ 5+ 3) = 13<br />
5<br />
1 2 3<br />
1 2 2 2 3 2<br />
2 2 2 4 2 8<br />
c = = = 2, c = = = 1, c = = ,<br />
1 1 2 4 3 9<br />
4 5<br />
2 16 2 32<br />
c4 = = = 1, c<br />
2 5 = =<br />
2<br />
4 16 5 25<br />
1 2 3 4 5<br />
1 = e = , 2 = e , 3 = e , e 4 = ,<br />
e<br />
5 =<br />
d e d d d d<br />
1 2 3 4 5<br />
2 2 4<br />
a = 3, a = ⋅ 3 = 2, a = ⋅ 2 = ,<br />
3 3 3<br />
2 4 8 2 8 16<br />
a4 = ⋅ = , a5<br />
= ⋅ =<br />
3 3 9 3 9 27<br />
5. 1 2 3<br />
1 1 1<br />
= 4, =− ⋅ 4 =− 1, =− ⋅− 1 = ,<br />
4 4 4<br />
1 1 1 1 1 1<br />
a4 =− ⋅ =− , a5<br />
=− ⋅− =<br />
4 4 16 4 16 64<br />
6. a1 a2 a3<br />
a = 2, a = 2 − 2 = 0, a = 2 − 0 = 2,<br />
7. 1 2 3<br />
a<br />
= 2− 2= 0, a = 2− 0=<br />
2<br />
4 5<br />
8. a1 a2 a3<br />
9.<br />
a<br />
= − 3, = 4 + ( − 3) = 1, = 4 + 1 = 5,<br />
= 4+ 5= 9, a = 4+ 9=<br />
13<br />
4 5<br />
4<br />
∑<br />
k = 1<br />
(4k<br />
+ 2)<br />
( 4 1 2) ( 4 2 2) ( 4 3 2) ( 4 4 2)<br />
( 6) ( 10) ( 14) ( 18)<br />
= ⋅ + + ⋅ + + ⋅ + + ⋅ +<br />
= + + +<br />
= 48<br />
3<br />
2 2 2 2<br />
10. ∑ (3 − k ) = ( 3 − 1 ) + ( 3 − 2 ) + ( 3 −3<br />
)<br />
k = 1<br />
( 2) ( 1) ( 6)<br />
= + − + −<br />
=−5<br />
1 1 1 1 k + 1 ⎛1⎞<br />
− + − +⋅⋅⋅+ = − ⎜ ⎟<br />
2 3 4 13<br />
⎝k<br />
⎠<br />
11. 1 ∑ ( 1)<br />
13<br />
k = 1<br />
2 3 4 n+ 1 n 1<br />
2 2 2 2 ⎛<br />
k+<br />
2 ⎞<br />
<strong>12</strong>. 2 + + + +⋅⋅⋅+ =<br />
2 3<br />
n ∑<br />
k<br />
3 3 3 3 ⎜<br />
k = 0 3 ⎟<br />
⎝ ⎠<br />
n+<br />
1⎛<br />
k<br />
2 ⎞<br />
= ∑<br />
⎜ k −1<br />
k = 1 3 ⎟<br />
⎝ ⎠<br />
13. { a } { n 5}<br />
n<br />
= + Arithmetic<br />
d = ( n+ 1+ 5) − ( n+ 5) = n+ 6−n− 5=<br />
1<br />
n<br />
n<br />
Sn<br />
= [ 6+ n+ 5 ] = ( n+<br />
11)<br />
2 2<br />
14. { b } { 4n<br />
3}<br />
n<br />
= + Arithmetic<br />
d = (4( n+ 1) + 3) − (4n+<br />
3)<br />
= 4n+ 4+ 3−4n− 3=<br />
4<br />
n<br />
Sn<br />
= + +<br />
2<br />
n<br />
= (4n<br />
+ 10)<br />
2<br />
= n +<br />
[ 7 4n<br />
3]<br />
( 2n<br />
5)<br />
<strong>12</strong>73<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
3<br />
15. { cn} { 2n<br />
}<br />
= Examine <strong>the</strong> terms of <strong>the</strong><br />
sequence: 2, 16, 54, <strong>12</strong>8, 250, ...<br />
There is no common difference; <strong>the</strong>re is no<br />
common ratio; nei<strong>the</strong>r.<br />
2<br />
16. { dn} { 2n<br />
1}<br />
= − Examine <strong>the</strong> terms of <strong>the</strong><br />
sequence: 1, 7, 17, 31, 49, ...<br />
There is no common difference; <strong>the</strong>re is no<br />
common ratio; nei<strong>the</strong>r.<br />
3<br />
17. { } { 2 n<br />
n }<br />
s = Geometric<br />
3( n+ 1) 3n+<br />
3<br />
2 2 3n+ 3−3n<br />
3<br />
r = = = 2 = 2 = 8<br />
3n<br />
3n<br />
2 2<br />
⎛<br />
n<br />
n<br />
1−8 ⎞ ⎛1−8 ⎞ 8 n<br />
S n = 8 ⎜ = 8 = 8 −1<br />
1 8 ⎟ ⎜ 7 ⎟<br />
⎝ − ⎠ ⎝ − ⎠ 7<br />
2<br />
18. { } { 3 n<br />
n }<br />
u = Geometric<br />
2( n+ 1) 2n+<br />
2<br />
( )<br />
3 3 2n+ 2−2n<br />
2<br />
r = = = 3 = 3 = 9<br />
2n<br />
2n<br />
3 3<br />
⎛<br />
n<br />
n<br />
1−9 ⎞ ⎛1−9 ⎞ 9 n<br />
S n = 9 ⎜ = 9 = 9 −1<br />
1 9 ⎟ ⎜ 8 ⎟<br />
⎝ − ⎠ ⎝ − ⎠ 8<br />
( )<br />
19. 0, 4, 8, <strong>12</strong>, ... Arithmetic d = 4− 0=<br />
4<br />
n<br />
n<br />
Sn<br />
= ( 2(0) + ( n− 1)4 ) = ( 4( n− 1) ) = 2 n( n−<br />
1)<br />
2 2<br />
20. 1, –3, –7, –11, ... Arithmetic<br />
d =−3− 1=−<br />
4<br />
21.<br />
n<br />
n<br />
Sn<br />
= + n− − =<br />
2 2<br />
− n+<br />
n<br />
= (6 − 4 n) = n( 3−2n)<br />
2<br />
( 2(1) ( 1)( 4) ) ( 2 4 4)<br />
3 3 3 3<br />
3, , , , , ...<br />
2 4 8 16<br />
⎛3<br />
⎞<br />
⎜ ⎟<br />
2 3 1 1<br />
r =<br />
⎝ ⎠<br />
= ⋅ =<br />
3 2 3 2<br />
S n<br />
Geometric<br />
⎛<br />
n<br />
n<br />
⎛1⎞ ⎞ ⎛ ⎛1⎞<br />
⎞<br />
⎜1−⎜ ⎟ ⎟ ⎜1−⎜ ⎟ ⎟<br />
n<br />
2 2 ⎛ ⎛1<br />
⎞ ⎞<br />
= 3⎜ ⎝ ⎠ ⎟= 3⎜ ⎝ ⎠ ⎟= 6 1−<br />
⎜ 1 ⎟ ⎜ 1<br />
⎜ ⎟<br />
1<br />
⎛ ⎞ ⎟ ⎜ ⎝2<br />
⎠ ⎟<br />
−<br />
⎝ ⎠<br />
⎜ 2 ⎟ ⎜ ⎜ ⎟<br />
⎝ 2 ⎠ ⎟<br />
⎝ ⎠ ⎝ ⎠<br />
22.<br />
5 5 5 5<br />
5, − , , − , , ... Geometric<br />
3 9 27 81<br />
⎛ 5 ⎞<br />
⎜−<br />
⎟<br />
3 5 1 1<br />
r =<br />
⎝ ⎠<br />
=− ⋅ =−<br />
5 3 5 3<br />
S<br />
n<br />
⎛<br />
n<br />
n<br />
⎛ 1⎞ ⎞ ⎛ ⎛ 1⎞<br />
⎞<br />
⎜1−⎜−<br />
⎟ ⎟ ⎜1−⎜−<br />
⎟ ⎟<br />
3 3<br />
= 5⎜ ⎝ ⎠ ⎟=<br />
5⎜ ⎝ ⎠ ⎟<br />
⎜ ⎛ 1⎞ ⎟ ⎜ ⎛4⎞<br />
⎟<br />
1− −<br />
⎜ ⎜ ⎟ 3 ⎟ ⎜ ⎜ ⎟<br />
3 ⎟<br />
⎝ ⎝ ⎠ ⎠ ⎝ ⎝ ⎠ ⎠<br />
n<br />
15 ⎛ ⎛ 1 ⎞ ⎞<br />
= 1− −<br />
4 ⎜ ⎜ ⎟<br />
⎝ 3⎠<br />
⎟<br />
⎝ ⎠<br />
23. Nei<strong>the</strong>r. There is no common difference or<br />
common ratio.<br />
24.<br />
25.<br />
26.<br />
27.<br />
28.<br />
3 5 7 9 11<br />
, , , , , Nei<strong>the</strong>r. There is no<br />
2 4 6 8 10<br />
common difference or common ratio.<br />
50 50<br />
k= 1 k=<br />
1<br />
( + )<br />
⎛50 50 1 ⎞<br />
3k<br />
= 3 k = 3⎜<br />
⎟=<br />
3825<br />
⎝ 2 ⎠<br />
∑ ∑<br />
30<br />
∑<br />
k = 1<br />
k<br />
2<br />
( + )( ⋅ + )<br />
30 30 1 2 30 1<br />
= = 9455<br />
6<br />
30 30 30 30 30<br />
∑ ∑ ∑ ∑ ∑<br />
(3k − 9) = 3k− 9 = 3 k−<br />
9<br />
k= 1 k= 1 k= 1 k= 1 k=<br />
1<br />
⎛30(30 + 1) ⎞<br />
= 3⎜<br />
⎟−30(9)<br />
⎝ 2 ⎠<br />
= 1395 − 270 = 1<strong>12</strong>5<br />
40 40 40<br />
∑ ∑ ∑<br />
( − 2k<br />
+ 8) = − 2k+<br />
8<br />
k= 1 k= 1 k=<br />
1<br />
40 40<br />
∑<br />
∑<br />
=− 2 k + 8<br />
k= 1 k=<br />
1<br />
⎛40(1 + 40) ⎞<br />
=− 2⎜<br />
⎟+<br />
40(8)<br />
⎝ 2 ⎠<br />
=− 1640 + 320 =−1320<br />
<strong>12</strong>74<br />
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.
<strong>Chapter</strong> <strong>12</strong> Review Exercises<br />
29.<br />
7<br />
∑<br />
k = 1<br />
10<br />
30. ∑ ( )<br />
k = 1<br />
⎛<br />
7<br />
⎞ ⎛<br />
7<br />
⎞<br />
⎛1⎞ ⎛1⎞<br />
k ⎜1−<br />
1<br />
1 1<br />
⎜ ⎟ ⎟ ⎜ −<br />
3 1<br />
⎜ ⎟ ⎟<br />
⎛ ⎞ ⎜ ⎝ ⎠ ⎟ ⎜ ⎝3⎠<br />
⎟<br />
⎜ ⎟ = =<br />
⎝3⎠ 3⎜ 1 ⎟ 3⎜ 2 ⎟<br />
1−<br />
⎛ ⎞<br />
⎜ 3 ⎟ ⎜ ⎜ ⎟<br />
3 ⎟<br />
⎝ ⎠ ⎝ ⎝ ⎠ ⎠<br />
1⎛<br />
1 ⎞<br />
= ⎜1−<br />
⎟<br />
2 ⎝ 2187 ⎠<br />
1 2186 1093<br />
= ⋅ = ≈0.49977<br />
2 2187 2187<br />
( )<br />
⎛<br />
10<br />
k 1− −2 ⎞<br />
⎛1−1024<br />
⎞<br />
− 2 = − 2⎜<br />
⎟=−2<br />
⎜<br />
⎜ ⎟<br />
1 −( −2) ⎟ 3<br />
⎝ ⎠<br />
⎝ ⎠<br />
2<br />
=− ( − 1023)<br />
= 682<br />
3<br />
31. Arithmetic<br />
a = 3, d = 4, a = a + ( n−<br />
1) d<br />
1 n 1<br />
a 9 = 3 + (9 − 1)4 = 3+ 8(4) = 3+ 32 = 35<br />
32. Arithmetic<br />
a = 1, d = − 2, a = a + ( n−1)<br />
d<br />
1 n 1<br />
a 8 = 1 + (8−1)( − 2) = 1+ 7( − 2) = 1− 14 = −13<br />
33. Geometric<br />
1<br />
n 1<br />
a1 = 1, r = , n = 11; an<br />
= a1r −<br />
10<br />
a<br />
11<br />
11−1 10<br />
⎛ 1 ⎞ ⎛ 1 ⎞<br />
= 1⋅ ⎜ ⎟ = ⎜ ⎟<br />
⎝ 10 ⎠ ⎝ 10 ⎠<br />
1<br />
=<br />
10,000,000,000<br />
34. Geometric<br />
n<br />
a = 1, r = 2, n= 11; a = a r −<br />
1 n 1<br />
11−1 10<br />
( ) ( )<br />
a 11 = 1⋅ 2 = 2 = 1024<br />
35. Arithmetic<br />
a = 2, d = 2, n = 9, a = a + ( n−<br />
1) d<br />
1 n 1<br />
a 9 = 2 + (9− 1) 2 = 2+<br />
8 2<br />
= 9 2 ≈<strong>12</strong>.7279<br />
36. Geometric<br />
n<br />
a = 2, r = 2, n= 9, a = a r −<br />
1 n 1<br />
−<br />
( ) ( )<br />
9 1 8<br />
a9 = 2 2 = 2 2 = 2⋅16<br />
= 16 2 ≈ 22.6274<br />
1<br />
1<br />
37. 7 1 20 1<br />
a = a + 6d = 31 a = a + 19d<br />
= 96;<br />
Solve <strong>the</strong> system of equations:<br />
a + 6d<br />
= 31<br />
1<br />
a1<br />
+ 19d<br />
= 96<br />
Subtract <strong>the</strong> second equation from <strong>the</strong> first<br />
equation and solve for d.<br />
− 13d<br />
=−65<br />
d = 5<br />
a 1 = 31− 6(5) = 31− 30 = 1<br />
a = a + n−<br />
d<br />
n<br />
( )<br />
( n )( )<br />
1 1<br />
= 1+ −1 5<br />
= 1+ 5n<br />
−5<br />
= 5n<br />
−4<br />
General formula: { a } = { 5n−<br />
4}<br />
38. 8 1 17 1<br />
n<br />
a = a + 7d =− 20 a = a + 16d<br />
= − 47;<br />
Solve <strong>the</strong> system of equations:<br />
a + 7d<br />
=−20<br />
1<br />
a1<br />
+ 16d<br />
=−47<br />
Subtract <strong>the</strong> second equation from <strong>the</strong> first<br />
equation and solve for d.<br />
− 9d<br />
= 27<br />
d = −3<br />
a 1 = −20 −7( − 3) = − 20 + 21 = 1<br />
a = a + n−<br />
d<br />
n<br />
( )<br />
( n )( )<br />
1 1<br />
= 1+ −1 −3<br />
= 1− 3n<br />
+ 3<br />
=− 3n<br />
+ 4<br />
General formula: { a } = { − 3n+<br />
4}<br />
39. 10 1 18 1<br />
a = a + 9d = 0 a = a + 17d<br />
= 8;<br />
Solve <strong>the</strong> system of equations:<br />
a + 9d<br />
= 0<br />
1<br />
a1<br />
+ 17d<br />
= 8<br />
Subtract <strong>the</strong> second equation from <strong>the</strong> first<br />
equation and solve for d.<br />
− 8d<br />
=−8<br />
d = 1<br />
a 1 = − 9(1) = − 9<br />
a = a + n−<br />
d<br />
n<br />
( )<br />
( n )( )<br />
1 1<br />
=− 9+ −1 1<br />
=− 9+ n −1<br />
= n −10<br />
General formula: { a } = { n−<br />
10}<br />
n<br />
n<br />
<strong>12</strong>75<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
40. <strong>12</strong> 1 22 1<br />
a = a + 11d = 30 a = a + 21d<br />
= 50 ;<br />
Solve <strong>the</strong> system of equations:<br />
a1<br />
+ 11d<br />
= 30<br />
a1<br />
+ 21d<br />
= 50<br />
Subtract <strong>the</strong> second equation from <strong>the</strong> first<br />
equation and solve for d.<br />
− 10d<br />
= −20<br />
d = 2<br />
a 1 = 30 − 11(2) = 30 − 22 = 8<br />
an<br />
= a1 + ( n−1)<br />
d<br />
= 8+ ( n −1)( 2)<br />
= 8+ 2n<br />
−2<br />
= 2n<br />
+ 6<br />
a = 2n+<br />
6<br />
General formula: { } { }<br />
1<br />
41. a1<br />
= 3, r =<br />
3<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
3 3 9<br />
Sn<br />
= = = =<br />
1− r ⎛ 1⎞ ⎛2⎞<br />
2<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 3 ⎠ ⎝ 3 ⎠<br />
1<br />
42. a1<br />
= 2, r =<br />
2<br />
Since r < 1, <strong>the</strong> series converges.<br />
a1<br />
2 2<br />
Sn<br />
= = = = 4<br />
1− r ⎛ 1⎞ ⎛1⎞<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 2 ⎠ ⎝ 2 ⎠<br />
1<br />
43. a1<br />
= 2, r = −<br />
2<br />
Since r < 1, <strong>the</strong> series converges.<br />
S<br />
n<br />
a1 2 2 4<br />
= = = =<br />
1− r ⎛ ⎛ 1 ⎞⎞ ⎛3<br />
⎞ 3<br />
⎜1−⎜− ⎟ ⎜ ⎟<br />
2<br />
⎟<br />
⎝ ⎝ ⎠⎠<br />
⎝2<br />
⎠<br />
2<br />
44. a1<br />
= 6, r = −<br />
3<br />
Since r < 1, <strong>the</strong> series converges.<br />
S<br />
n<br />
a1 6 6 18<br />
= = = =<br />
1− r ⎛ ⎛ 2 ⎞⎞ ⎛5<br />
⎞ 5<br />
⎜1−⎜− ⎟ ⎜ ⎟<br />
3<br />
⎟<br />
⎝ ⎝ ⎠⎠<br />
⎝3<br />
⎠<br />
n<br />
1 3<br />
45. a<br />
1<br />
= , r =<br />
2 2<br />
Since r > 1, <strong>the</strong> series diverges.<br />
5<br />
46. a 1 = 5 , r = −<br />
4<br />
Since r > 1, <strong>the</strong> series diverges.<br />
1<br />
47. a1<br />
= 4, r =<br />
2<br />
Since r < 1, <strong>the</strong> series converges.<br />
S<br />
n<br />
a1 4 4<br />
= = = = 8<br />
1− r ⎛ 1⎞ ⎛1⎞<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 2 ⎠ ⎝ 2 ⎠<br />
3<br />
48. a1<br />
= 3, r = −<br />
4<br />
Since r < 1, <strong>the</strong> series converges.<br />
S<br />
49. I:<br />
50. I:<br />
n<br />
a1 3 3 <strong>12</strong><br />
= = = =<br />
1− r ⎛ ⎛ 3 ⎞⎞ ⎛7<br />
⎞ 7<br />
⎜1−⎜− ⎟ ⎜ ⎟<br />
4<br />
⎟<br />
⎝ ⎝ ⎠⎠<br />
⎝4<br />
⎠<br />
31 ⋅<br />
n = 1: 3⋅ 1= 3and (1+ 1) = 3<br />
2<br />
3k<br />
II: If 3+ 6+ 9+ + 3 k = ( k+<br />
1) , <strong>the</strong>n<br />
2<br />
3+ 6+ 9+ + 3k+ 3( k+<br />
1)<br />
= [ 3+ 6+ 9+ + 3k]<br />
+ 3( k+<br />
1)<br />
3 k<br />
= ( k + 1) + 3( k + 1)<br />
2<br />
⎛3k<br />
⎞ 3( k + 1)<br />
= ( k + 1) ⎜ + 3 ⎟= [( k+ 1) + 1]<br />
⎝ 2 ⎠ 2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
2<br />
n = 1: 4⋅1− 2 = 2 and 2(1) = 2<br />
II: If 2+ 6+ 10 + + (4k− 2) = 2k<br />
, <strong>the</strong>n<br />
2 + 6 + 10 + + (4k− 2) + [4( k+ 1) −2]<br />
= 2 + 6 + 10 + + (4k− 2) + 4k+<br />
2<br />
[ ]<br />
2 2<br />
( ) ( ) 2<br />
= 2k + 4k+ 2= 2 k + 2k+ 1 = 2 k+<br />
1<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
2<br />
<strong>12</strong>76<br />
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<strong>Chapter</strong> <strong>12</strong> Review Exercises<br />
51. I:<br />
11 −<br />
1<br />
n = 1: 2⋅ 3 = 2 and 3 − 1=<br />
2<br />
k−1<br />
II: If 2+ 6+ 18+ + 2⋅ 3 = 3 −1, <strong>the</strong>n<br />
k−<br />
1 k+ 1−1<br />
2+ 6+ 18+ + 2⋅ 3 + 2⋅3<br />
k−1<br />
k<br />
= ⎡2 6 18 23 ⎤<br />
⎣<br />
+ + + + ⋅<br />
⎦<br />
+ 23 ⋅<br />
k k k k+<br />
1<br />
= 3 − 1+ 2⋅ 3 = 3⋅3 − 1= 3 −1<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
11 −<br />
1<br />
52. I: n = 1: 3⋅ 2 = 3and 3( 2 − 1)<br />
= 3<br />
k−1<br />
k<br />
II: If 3+ 6+ <strong>12</strong>+ + 3⋅ 2 = 3( 2 −1)<br />
k<br />
k−<br />
1 k+ 1−1<br />
3+ 6+ <strong>12</strong>+ + 32 ⋅ + 32 ⋅<br />
k−1<br />
= ⎡3 6 <strong>12</strong> 32 ⎤<br />
⎣<br />
+ + + + ⋅<br />
⎦<br />
+ 32 ⋅<br />
= 32 − 1+ 32 ⋅<br />
, <strong>the</strong>n<br />
k<br />
k<br />
( )<br />
k k k k+<br />
1<br />
( ) ( ) ( )<br />
= 3⋅ 2 − 1+ 2 = 3 2⋅2 − 1 = 3 2 −1<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
k<br />
54. I:<br />
1 3 2 2<br />
= ⋅ ⎡6k 6k 9k 9k 2k<br />
2⎤<br />
2 ⎣<br />
+ + + + +<br />
⎦<br />
1 2<br />
= ⋅ ⎡6k ( k+ 1) + 9k( k+ 1) + 2( k+<br />
1)<br />
⎤<br />
2 ⎣<br />
⎦<br />
1 ( 1) 6 2<br />
k ⎡ k <strong>12</strong> k 6 3 k 3 1<br />
= ⋅ + + + − − − ⎤<br />
2 ⎣<br />
⎦<br />
1 2<br />
= ⋅ ( k+ 1) ⎡ 6( k 2 k 1) 3( k 1) 1 ⎤<br />
2 ⎣<br />
+ + − + −<br />
⎦<br />
1 ( 1) 6( 1) 2<br />
= ⋅ k+ ⎡ k+ − 3( k+ 1) − 1 ⎤<br />
2 ⎣<br />
⎦<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
1<br />
n = 1: 1(1 + 2) = 3and ⋅ (1+ 1)(2⋅ 1+ 7) = 3<br />
6<br />
II: If<br />
k<br />
13 ⋅ + 24 ⋅ + + kk ( + 2) = ( k+ 1)(2k+<br />
7) ,<br />
6<br />
<strong>the</strong>n<br />
13 ⋅ + 24 ⋅ + + kk ( + 2) + ( k+ 1)( k+ 1+<br />
2)<br />
= 13 ⋅ + 24 ⋅ + + kk ( + 2) + ( k+ 1)( k+<br />
3)<br />
[ ]<br />
53. I: n = 1:<br />
k<br />
= ( k + 1)(2 k + 7) + ( k + 1)( k + 3)<br />
2 1<br />
6<br />
2<br />
(3⋅1− 2) = 1 and ⋅1(6 ⋅1 −3⋅1− 1) = 1<br />
( k + 1)<br />
2<br />
2<br />
= ( 2 k + 7 k+ 6 k+<br />
18 )<br />
6<br />
II: If<br />
( k + 1) 2<br />
2 2 2 1 2<br />
=<br />
1 + 4 + + (3k− 2) = ⋅k( 6k −3k−1)<br />
,<br />
( 2 k + 13 k+<br />
18 )<br />
6<br />
2<br />
<strong>the</strong>n<br />
( k + 1)<br />
= [( k + 1) + 1][2( k+ 1) + 7]<br />
2 2 2 2<br />
2<br />
1 + 4 + 7 + + (3k− 2) + ( 3( k+ 1) −2<br />
6<br />
)<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
⎡<br />
2 2 2 2 2<br />
= 1 + 4 + 7 + + (3k− 2) ⎤<br />
⎣<br />
<br />
⎦<br />
+ (3k+<br />
1)<br />
true.<br />
1 2 2<br />
= ⋅k( 6k −3k− 1 ) + (3k+<br />
1)<br />
⎛5⎞ 5! 54321 ⋅ ⋅ ⋅ ⋅ 54 ⋅<br />
2<br />
55. ⎜ ⎟= = = = 10<br />
1<br />
⎝2⎠<br />
2!3! 21321 ⋅ ⋅ ⋅ ⋅ 21 ⋅<br />
⎡<br />
3 2 2<br />
= ⋅ 6k −3k − k+ 18k + <strong>12</strong>k+<br />
2⎤<br />
2 ⎣<br />
⎦<br />
⎛8⎞ 1<br />
8! 87654321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 87 ⋅<br />
⎡<br />
3 2<br />
= ⋅ 6k + 15k + 11k+<br />
2⎤<br />
56. ⎜ ⎟= = = = 28<br />
2 ⎣<br />
⎦<br />
⎝6⎠<br />
6!2! 65432<strong>12</strong>1 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 21 ⋅<br />
1 ( 1) 6 2<br />
= ⋅ k+ ⎡ k + 9 k+<br />
2 ⎤<br />
2 ⎣ ⎦<br />
_________________________________________________________________________________________________<br />
57.<br />
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞<br />
( x+ 2) = ⎜ ⎟x + ⎜ ⎟x ⋅ 2+ ⎜ ⎟x ⋅ 2 + ⎜ ⎟x ⋅ 2 + ⎜ ⎟x<br />
⋅ 2 + ⎜ ⎟⋅2<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠<br />
5 5 4 3 2 2 3 1 4 5<br />
5 4 3 2<br />
= x + 5⋅ 2x + 10⋅ 4x + 10⋅ 8x + 5⋅ 16x+ 1⋅32<br />
5 4 3 2<br />
= x + 10x + 40x + 80x + 80x+<br />
32<br />
<strong>12</strong>77<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
58.<br />
59.<br />
60.<br />
⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞<br />
( x− 3) = ⎜ ⎟x + ⎜ ⎟x ( − 3) + ⎜ ⎟x ( − 3) + ⎜ ⎟x( − 3) + ⎜ ⎟x<br />
( −3)<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠<br />
4 4 3 2 2 3 0 4<br />
4 3 2<br />
= x + 4( − 3) x + 6⋅ 9x + 4( − 27) x+<br />
81<br />
4 3 2<br />
= x − <strong>12</strong>x + 54x − 108x+<br />
81<br />
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞<br />
(2x+ 3) = ⎜ ⎟(2 x) + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟⋅3<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠<br />
5 5 4 3 2 2 3 1 4 5<br />
5 4 3 2<br />
= 32x + 5⋅16x ⋅ 3 + 10⋅8x ⋅ 9 + 10⋅4x ⋅ 27 + 5⋅2x⋅ 81+ 1⋅243<br />
5 4 3 2<br />
= 32x + 240x + 720x + 1080x + 810x+<br />
243<br />
⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞<br />
(3x− 4) = ⎜ ⎟(3 x) + ⎜ ⎟(3 x) ( − 4) + ⎜ ⎟(3 x) ( − 4) + ⎜ ⎟(3 x)( − 4) + ⎜ ⎟( −4)<br />
⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠<br />
4 4 3 2 2 3 4<br />
4 3 2<br />
= 81x + 4⋅27 x ( − 4) + 6⋅9x ⋅ 16 + 4⋅3 x( − 64) + 1⋅256<br />
4 3 2<br />
= 81x − 432x + 864x − 768x+<br />
256<br />
_________________________________________________________________________________________________<br />
61. n = 9, j = 2, x = x, a = 2<br />
⎛9⎞<br />
9! 9⋅8<br />
⎜ ⎟x ⋅ 2 = ⋅ 4x = ⋅ 4x = 144x<br />
⎝2⎠<br />
2!7! 2⋅1<br />
7<br />
The coefficient of x is 144.<br />
7 2 7 7 7<br />
62. n = 8, j = 5, x = x, a =− 3<br />
⎛8⎞<br />
3 5 8!<br />
3<br />
⎜ ⎟x<br />
( − 3) = ( −243)<br />
x<br />
⎝5⎠<br />
5!3!<br />
876 ⋅ ⋅<br />
= ( − 243) x =− 13,608 x<br />
321 ⋅ ⋅<br />
3<br />
The coefficient of x is − 13,608.<br />
63. n = 7, j = 5, x = 2 x, a = 1<br />
3 3<br />
⎛7⎞<br />
7! 7⋅6<br />
⎜ ⎟(2 x) ⋅ 1 = ⋅ 4 x (1) = ⋅ 4x = 84x<br />
⎝5⎠<br />
5!2! 2⋅1<br />
2<br />
The coefficient of x is 84.<br />
2 5 2 2 2<br />
64. n = 8, j = 2, x = 2 x, a = 1<br />
⎛8⎞<br />
6 2 8! 6<br />
⎜ ⎟(2 x) ⋅ 1 = ⋅64 x (1)<br />
⎝2⎠<br />
2!6!<br />
87 ⋅<br />
= ⋅ 64 x = 1792 x<br />
21 ⋅<br />
6<br />
The coefficient of x is 1792.<br />
6 6<br />
65. This is an arithmetic sequence with<br />
a1 = 80, d =− 3, n=<br />
25<br />
a. a 25 = 80 + (25 −1)( − 3) = 80 − 72 = 8 bricks<br />
25<br />
S = (80 + 8) = 25(44) = 1100 bricks<br />
2<br />
1100 bricks are needed to build <strong>the</strong> steps.<br />
b. 25<br />
66. This is an arithmetic sequence with<br />
a1 = 30, d =− 1, a n = 15<br />
15 = 30 + ( n −1)( −1)<br />
− 15 =− n + 1<br />
− 16 =−n<br />
n = 16<br />
16<br />
S 16 = (30 + 15) = 8(45) = 360 tiles<br />
2<br />
360 tiles are required to make <strong>the</strong> trapezoid.<br />
67. This is a geometric sequence with<br />
3<br />
a1<br />
= 20, r = .<br />
4<br />
a. After striking <strong>the</strong> ground <strong>the</strong> third time, <strong>the</strong><br />
3<br />
⎛3⎞ 135<br />
height is 20⎜<br />
⎟ = ≈ 8.44 feet .<br />
⎝4⎠<br />
16<br />
th<br />
b. After striking <strong>the</strong> ground <strong>the</strong> n time, <strong>the</strong><br />
n<br />
⎛3<br />
⎞<br />
height is 20 ⎜ ⎟ feet .<br />
⎝4<br />
⎠<br />
<strong>12</strong>78<br />
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<strong>Chapter</strong> <strong>12</strong> Review Exercises<br />
c. If <strong>the</strong> height is less than 6 inches or 0.5 feet,<br />
<strong>the</strong>n:<br />
n<br />
⎛3<br />
⎞<br />
0.5 ≥ 20⎜ ⎟<br />
⎝ 4 ⎠<br />
n<br />
⎛3<br />
⎞<br />
0.025 ≥ ⎜ ⎟<br />
⎝4<br />
⎠<br />
⎛3<br />
⎞<br />
log ( 0.025)<br />
≥ nlog ⎜ ⎟<br />
⎝ 4 ⎠<br />
log ( 0.025)<br />
n ≥<br />
≈<strong>12</strong>.82<br />
⎛3<br />
⎞<br />
log ⎜ ⎟<br />
⎝ 4 ⎠<br />
The height is less than 6 inches after <strong>the</strong><br />
13th strike.<br />
d. Since this is a geometric sequence with<br />
r < 1 , <strong>the</strong> distance is <strong>the</strong> sum of <strong>the</strong> two<br />
infinite geometric series - <strong>the</strong> distances<br />
going down plus <strong>the</strong> distances going up.<br />
Distance going down:<br />
20 20<br />
S down = = = 80 feet.<br />
⎛ 3⎞ ⎛1⎞<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 4 ⎠ ⎝ 4 ⎠<br />
Distance going up:<br />
15 15<br />
S up = = = 60 feet.<br />
⎛ 3⎞ ⎛1⎞<br />
⎜1−<br />
⎟ ⎜ ⎟<br />
⎝ 4 ⎠ ⎝ 4 ⎠<br />
The total distance traveled is 140 feet.<br />
68. a. Since <strong>the</strong> interest rate is 6.75% per annum<br />
compounded monthly, this is equivalent to a<br />
rate of (6.75/<strong>12</strong>)% each month. Defining a<br />
recursive sequence, we have:<br />
A0<br />
= 190,000<br />
⎛ 0.0675 ⎞<br />
An<br />
= ⎜1+ ⎟An<br />
− 1 −<strong>12</strong>32.34<br />
⎝ <strong>12</strong> ⎠<br />
⎛ 0.0675 ⎞<br />
⎜1+ ⎟ 190,000 −<strong>12</strong>32.34<br />
⎝ <strong>12</strong> ⎠<br />
= $189,836.41<br />
b. ( )<br />
c. Enter <strong>the</strong> recursive formula in Y= and create<br />
<strong>the</strong> table:<br />
d. Scroll through <strong>the</strong> table:<br />
After 252 months, <strong>the</strong> balance is below<br />
$100,000. The balance is about $99,540.<br />
e. Scroll through <strong>the</strong> table:<br />
The loan will be paid off after 360 months (or<br />
30 years). They will make 359 payments of<br />
$<strong>12</strong>32.34 plus a last payment of $<strong>12</strong>21.27 plus<br />
interest. The total amount paid is:<br />
⎛ 0.0675 ⎞<br />
359(<strong>12</strong>32.34) + <strong>12</strong>21.27⎜1+<br />
⎟<br />
⎝ <strong>12</strong> ⎠<br />
≈ $443,638.20<br />
f. The total interest expense is <strong>the</strong> difference<br />
of <strong>the</strong> total of <strong>the</strong> payments and <strong>the</strong> original<br />
loan: 443,638.20 − 190,000 = $253,638.20 .<br />
g. (a) Since <strong>the</strong> interest rate is 6.75% per annum<br />
compounded monthly, this is equivalent to<br />
a rate of (6.75/<strong>12</strong>)% each month.<br />
Defining a recursive sequence, we have:<br />
A0<br />
= 190,000<br />
⎛ 0.0675 ⎞<br />
An<br />
= ⎜1+ ⎟An<br />
− 1 −1332.34<br />
⎝ <strong>12</strong> ⎠<br />
⎛ 0.0675 ⎞<br />
⎜1+ ⎟ 190,000 −1332.34<br />
⎝ <strong>12</strong> ⎠<br />
= $189,736.41<br />
(b) ( )<br />
(c) Enter <strong>the</strong> recursive formula in Y= and<br />
create <strong>the</strong> table:<br />
<strong>12</strong>79<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
c. Scroll through <strong>the</strong> table:<br />
(d) Scroll through <strong>the</strong> table:<br />
At <strong>the</strong> beginning of <strong>the</strong> 33rd month, or after<br />
32 payments have been made, <strong>the</strong> balance is<br />
below $4,000. The balance is $3982.8.<br />
d. Scroll through <strong>the</strong> table:<br />
At <strong>the</strong> beginning of <strong>the</strong> 192nd month, or<br />
after 191 payments have been made, <strong>the</strong><br />
balance is below $100,000. The<br />
balance is about $99,288.<br />
(e) Scroll through <strong>the</strong> table:<br />
The loan will be paid off after 289<br />
months. They will make 288 payments<br />
of $1332.34 plus a last payment of<br />
$1142.80 plus interest. The total<br />
amount paid is:<br />
⎛ 0.0675 ⎞<br />
288(1332.34) + 1142.80⎜1+<br />
⎟<br />
⎝ <strong>12</strong> ⎠<br />
≈ $384,863.15<br />
(f) The total interest expense is <strong>the</strong><br />
difference of <strong>the</strong> total of <strong>the</strong> payments<br />
and <strong>the</strong> original loan:<br />
384,863.15 − 190,000 = $19,863.15 .<br />
69. a. b1 = 5,000, bn<br />
= 1.015bn<br />
− 1−<br />
100<br />
b2 = 1.015b( 2−1)<br />
− 100 = 1.015b1−100<br />
= 1.015 5000 − 100 = $4975<br />
( )<br />
b. Enter <strong>the</strong> recursive formula in Y= and draw<br />
<strong>the</strong> graph:<br />
The balance will be paid off at <strong>the</strong> end of 94<br />
months or 7years and 10 months.<br />
Total payments =<br />
100(93) + 11.01 1.015 = $9,311.18<br />
( )<br />
e. The total interest expense is <strong>the</strong> difference<br />
between <strong>the</strong> total of <strong>the</strong> payments and <strong>the</strong><br />
original balance:<br />
100(93) + 11.18 − 5,000 = $4,311.18<br />
70. This is a geometric sequence with<br />
a1 = 20,000, r = 1.04, n = 5 . Find <strong>the</strong> fifth term<br />
of <strong>the</strong> sequence:<br />
5−1 4<br />
a 5 = 20,000(1.04) = 20,000(1.04) = 23,397.17<br />
Her salary in <strong>the</strong> fifth year will be $23,397.17.<br />
<strong>Chapter</strong> <strong>12</strong> Test<br />
1.<br />
2<br />
n −1<br />
an<br />
=<br />
n + 8<br />
2 2<br />
1 −1 0 2 −1 3<br />
a1 = = = 0, a2<br />
= = ,<br />
1+ 8 9 2+<br />
8 10<br />
2 2<br />
3 −1 8 4 −1 15 5<br />
a3 = = , a4<br />
= = = ,<br />
3+ 8 11 4+<br />
8 <strong>12</strong> 4<br />
2<br />
5 −1 24<br />
a5<br />
= =<br />
5+<br />
8 13<br />
The first five terms of <strong>the</strong> sequence are 0, 3<br />
10 ,<br />
8<br />
11 , 5 24<br />
, and<br />
4 13 .<br />
<strong>12</strong>80<br />
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<strong>Chapter</strong> <strong>12</strong> Test<br />
2. a1 an<br />
an<br />
− 1<br />
= 4; = 3 + 2<br />
a2 = 3a1+ 2= 3( 4)<br />
+ 2=<br />
14<br />
a3 = 3a2<br />
+ 2= 3( 14)<br />
+ 2=<br />
44<br />
a4 = 3a3<br />
+ 2= 3( 44)<br />
+ 2=<br />
134<br />
a = 3a<br />
+ 2 = 3 134 + 2 = 404<br />
5 4<br />
( )<br />
The first five terms of <strong>the</strong> sequence are 4, 14, 44,<br />
134, and 404.<br />
3<br />
3. ∑ ( 1)<br />
4.<br />
5.<br />
k = 1<br />
k + 1 ⎛k<br />
+ 1⎞<br />
− ⎜ 2 ⎟<br />
⎝ k ⎠<br />
11 + 1+ 1 21 + 2+ 1 31 + 3+<br />
1<br />
= ( − 1<br />
⎛ ⎞<br />
) ( 1<br />
⎛ ⎞<br />
) ( 1<br />
⎛ ⎞<br />
⎜ )<br />
2<br />
⎟+ − ⎜<br />
2<br />
⎟+ − ⎜ 2 ⎟<br />
⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠<br />
2⎛2⎞ 3⎛3⎞ 4⎛4⎞<br />
= ( − 1) ⎜ ( 1) ( 1)<br />
1<br />
⎟+ − ⎜<br />
4<br />
⎟+ − ⎜<br />
9<br />
⎟<br />
⎝ ⎠ ⎝ ⎠ ⎝ ⎠<br />
3 4 61<br />
= 2 − + = 4 9 36<br />
4<br />
∑<br />
k<br />
⎡⎛2<br />
⎞ ⎤<br />
⎢⎜<br />
− k ⎥<br />
3<br />
⎟<br />
⎢⎣⎝<br />
⎠ ⎥⎦<br />
k = 1<br />
⎡ 1 ⎤ ⎡ 2 ⎤ ⎡ 3 ⎤ ⎡ 4<br />
⎤<br />
2 2 2 2<br />
( ) ( ) ( ) ( )<br />
= ⎢ − 1 2 3 4<br />
3 ⎥ + ⎢ −<br />
3 ⎥ + ⎢ −<br />
3 ⎥ + ⎢ −<br />
3 ⎥<br />
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦<br />
⎛2 ⎞ ⎛4 ⎞ ⎛ 8 ⎞ ⎛16<br />
⎞<br />
= ⎜ − 1 2 3 4<br />
3<br />
⎟+ ⎜ −<br />
9<br />
⎟+ ⎜ −<br />
27<br />
⎟+ ⎜ −<br />
81<br />
⎟<br />
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠<br />
1 14 73 308<br />
=− − − −<br />
3 9 27 81<br />
680<br />
=−<br />
81<br />
2 3 4 11<br />
− + − + ... +<br />
5 6 7 14<br />
Notice that <strong>the</strong> signs of each term alternate, with<br />
<strong>the</strong> first term being negative. This implies that<br />
<strong>the</strong> general term will include a power of − 1 .<br />
Also note that <strong>the</strong> numerator is always 1 more<br />
than <strong>the</strong> term number and <strong>the</strong> denominator is 4<br />
more than <strong>the</strong> term number. Thus, each term is in<br />
k ⎛ k + 1 ⎞<br />
<strong>the</strong> form ( −1)<br />
⎜<br />
k + 4<br />
⎟. The last numerator is 11<br />
⎝ ⎠<br />
which indicates that <strong>the</strong>re are 10 terms.<br />
10<br />
2 3 4 11 k ⎛ k + 1⎞<br />
− + − + ... + = ( 1)<br />
5 6 7 14<br />
∑ − ⎜<br />
k + 4<br />
⎟<br />
⎝ ⎠<br />
k = 1<br />
6. 6,<strong>12</strong>,36,144,...<br />
<strong>12</strong> − 6 = 6 and 36 − <strong>12</strong> = 24<br />
The difference between consecutive terms is not<br />
constant. Therefore, <strong>the</strong> sequence is not<br />
arithmetic.<br />
7.<br />
<strong>12</strong> = 2 and 36<br />
6 <strong>12</strong><br />
= 3<br />
The ratio of consecutive terms is not constant.<br />
Therefore, <strong>the</strong> sequence is not geometric.<br />
1 4<br />
n<br />
a n = − ⋅<br />
2<br />
1 n 1 n−1<br />
a − ⋅4 − ⋅4 ⋅4<br />
n 2 2<br />
= = = 4<br />
a 1 n−1 1 n−1<br />
− ⋅4 − ⋅4<br />
n−1 2 2<br />
Since <strong>the</strong> ratio of consecutive terms is constant,<br />
<strong>the</strong> sequence is geometric with common ratio<br />
1 1<br />
r = 4 and first term a 1 =− ⋅ 4 =− 2 .<br />
2<br />
The sum of <strong>the</strong> first n terms of <strong>the</strong> sequence is<br />
given by<br />
n<br />
1−<br />
r<br />
Sn<br />
= a1<br />
⋅<br />
1−<br />
r<br />
n<br />
1−<br />
4<br />
=−2⋅<br />
1 − 4<br />
2<br />
( 1 4<br />
n<br />
= − )<br />
3<br />
8. −2, −10, −18, − 26,...<br />
−10 −( − 2)<br />
= − 8 , ( )<br />
−18 − − 10 = − 8 ,<br />
−26 −( − 18)<br />
= − 8<br />
The difference between consecutive terms is<br />
constant. Therefore, <strong>the</strong> sequence is arithmetic<br />
with common difference d =− 8 and first term<br />
a 1 = − 2 .<br />
an<br />
= a1 + ( n−1)<br />
d<br />
= − 2+ ( n −1)( −8)<br />
=−2− 8n<br />
+ 8<br />
= 6−8n<br />
The sum of <strong>the</strong> first n terms of <strong>the</strong> sequence is<br />
given by<br />
n<br />
Sn<br />
= ( a+<br />
an)<br />
2<br />
n<br />
= ( − 2+ 6−8n)<br />
2<br />
n<br />
= ( 4−8n)<br />
2<br />
= n 2−4n<br />
( )<br />
<strong>12</strong>81<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
n<br />
9. a n =− + 7<br />
2<br />
⎡ n ⎤ ⎡ ( n −1)<br />
⎤<br />
an<br />
− an−1<br />
=<br />
⎢<br />
− + 7 − 7<br />
2<br />
⎢− +<br />
2<br />
⎥<br />
⎣<br />
⎥<br />
⎦ ⎣ ⎦<br />
n n−1<br />
=− + 7+ −7<br />
2 2<br />
1<br />
=−<br />
2<br />
The difference between consecutive terms is<br />
constant. Therefore, <strong>the</strong> sequence is arithmetic<br />
1<br />
with common difference d =− and first term<br />
2<br />
1 13<br />
a 1 =− + 7 = .<br />
2 2<br />
The sum of <strong>the</strong> first n terms of <strong>the</strong> sequence is<br />
given by<br />
n<br />
Sn<br />
= ( a1<br />
+ an)<br />
2<br />
n⎛13<br />
⎛ n ⎞⎞<br />
= 7<br />
2⎜<br />
+<br />
2<br />
⎜− +<br />
2<br />
⎟⎟<br />
⎝ ⎝ ⎠⎠<br />
n⎛27<br />
n⎞<br />
=<br />
2<br />
⎜ −<br />
2 2<br />
⎟<br />
⎝ ⎠<br />
n<br />
= ( 27 −n)<br />
4<br />
10.<br />
8<br />
25,10, 4, ,...<br />
5<br />
10 2<br />
= , 4 2<br />
8<br />
5 8 1 2<br />
= , = ⋅ =<br />
25 5 10 5 4 5 4 5<br />
The ratio of consecutive terms is constant.<br />
Therefore, <strong>the</strong> sequence is geometric with<br />
common ratio r = 2<br />
5<br />
and first term a 1 = 25 .<br />
The sum of <strong>the</strong> first n terms of <strong>the</strong> sequence is<br />
given by<br />
n<br />
2<br />
n<br />
1− ⎛ ⎞<br />
⎜<br />
1<br />
5<br />
⎟<br />
− r<br />
Sn<br />
= a1<br />
⋅ = 25⋅<br />
⎝ ⎠<br />
1−<br />
r 2<br />
1−<br />
5<br />
11.<br />
2n<br />
− 3<br />
an<br />
=<br />
2n<br />
+ 1<br />
2n<br />
− 3 2( n −1)<br />
−3<br />
an<br />
− an−1<br />
= −<br />
2n<br />
+ 1 2( n − 1)<br />
+ 1<br />
2n−3 2n−5<br />
= −<br />
2n+ 1 2n−1<br />
=<br />
( 2n− 3)( 2n−1) −( 2n− 5)( 2n+<br />
1)<br />
( 2n+ 1)( 2n−1)<br />
2 2<br />
( 4n − 8n+ 3) −( 4n −8n−5)<br />
=<br />
2<br />
4n<br />
−1<br />
8<br />
=<br />
2<br />
4n<br />
− 1<br />
The difference of consecutive terms is not<br />
constant. Therefore, <strong>the</strong> sequence is not<br />
arithmetic.<br />
2n<br />
− 3<br />
an<br />
=<br />
2n<br />
+ 1<br />
an−1<br />
2( n −1)<br />
−3<br />
2( n − 1)<br />
+ 1<br />
2n−3 2n−1<br />
= ⋅<br />
2n+ 1 2n−5<br />
( 2n−<br />
3)( 2n−1)<br />
=<br />
2n+ 1 2n−<br />
5<br />
( )( )<br />
The ratio of consecutive terms is not constant.<br />
Therefore, <strong>the</strong> sequence is not geometric.<br />
−64 1<br />
<strong>12</strong>. For this geometric series we have r = =−<br />
256 4<br />
1 1<br />
and a 1 = 256 . Since r =− = < 1, <strong>the</strong> series<br />
4 4<br />
converges and we get<br />
a1<br />
256 256 1024<br />
S∞ = 1−<br />
r<br />
= 1 5<br />
1−( −<br />
4 )<br />
= = 5<br />
4<br />
n<br />
⎡ 2<br />
1<br />
⎛ ⎞ ⎤<br />
⎢ − ⎜<br />
5<br />
⎟ ⎥<br />
n<br />
⎢ ⎝ ⎠ ⎥ 5⎡<br />
⎛2⎞<br />
⎤<br />
= 25⋅ ⎣ ⎦<br />
= 25⋅ ⎢1<br />
−<br />
3 3<br />
⎜<br />
5<br />
⎟ ⎥<br />
⎢⎣<br />
⎝ ⎠ ⎥⎦<br />
5<br />
n<br />
<strong>12</strong>5 ⎡ ⎛2<br />
⎞ ⎤<br />
= ⎢1<br />
−<br />
3<br />
⎜ ⎥<br />
5<br />
⎟<br />
⎢⎣<br />
⎝ ⎠ ⎥⎦<br />
<strong>12</strong>82<br />
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<strong>Chapter</strong> <strong>12</strong> Test<br />
⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞<br />
⎜0⎟ ⎜1⎟ ⎜2⎟ ⎜3⎟ ⎜4⎟ ⎜5⎟<br />
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠<br />
5 4 3 2<br />
= 243m + 5⋅81m ⋅ 2 + 10⋅27m ⋅ 4 + 10⋅9m ⋅ 8 + 5⋅3m⋅ 16 + 32<br />
5 4 3 2<br />
= 243m + 810m + 1080m + 720m + 240m+<br />
32<br />
5 5 4 3 2 2 3 4 5<br />
13. ( 3m+ 2) = ( 3m) + ( 3m) ( 2) + ( 3m) ( 2) + ( 3m) ( 2) + ( 3m)( 2) + ( 2)<br />
14. First we show that <strong>the</strong> statement holds for n = 1 .<br />
⎛ 1⎞<br />
⎜1+ = 1+ 1=<br />
2<br />
1<br />
⎟<br />
⎝ ⎠<br />
1 1 1 1<br />
The equality is true for n = 1 so Condition I holds. Next we assume that ⎜ ⎛ 1+ 1 1 ... 1 n 1<br />
1<br />
⎟⎜ ⎞⎛ + ⎞⎛ + ⎞ ⎛ + ⎞ = +<br />
2<br />
⎟⎜<br />
3<br />
⎟ ⎜<br />
n<br />
⎟<br />
⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠<br />
is true for some k, and we show <strong>the</strong> formula <strong>the</strong>n holds for k + 1. We assume that<br />
⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞<br />
⎜1+ 1 1 ... 1 k 1<br />
1<br />
⎟⎜ + + + = +<br />
2<br />
⎟⎜<br />
3<br />
⎟ ⎜<br />
k<br />
⎟ . Now we need to show that<br />
⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠<br />
⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎛ 1 ⎞<br />
⎜1+ 1 1 ... 1 1 ( k 1)<br />
1 k 2<br />
1<br />
⎟⎜ +<br />
2<br />
⎟⎜ + + + = + + = +<br />
3<br />
⎟ ⎜<br />
k<br />
⎟⎜<br />
k+<br />
1<br />
⎟<br />
.<br />
⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠<br />
We do this as follows:<br />
⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎛ 1 ⎞ ⎡⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎤⎛ 1 ⎞<br />
⎜1+ 1 1 ... 1 1 1 1 1 ... 1 1<br />
1<br />
⎟⎜ +<br />
2<br />
⎟⎜ +<br />
3<br />
⎟ ⎜ +<br />
k<br />
⎟⎜ +<br />
k 1<br />
⎟ = ⎜ +<br />
1<br />
⎟⎜ +<br />
2<br />
⎟⎜ +<br />
3<br />
⎟ ⎜ +<br />
k<br />
⎟ ⎜ +<br />
k 1<br />
⎟<br />
⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ + ⎠<br />
⎢<br />
⎥<br />
⎣⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎦⎝ + ⎠<br />
⎛ 1 ⎞<br />
= ( k + 1)<br />
⎜1 + (using <strong>the</strong> induction assumption)<br />
k 1<br />
⎟<br />
⎝ + ⎠<br />
1<br />
= ( k+ 1) ⋅ 1+ ( k+ 1)<br />
⋅<br />
k + 1<br />
= k + 1+<br />
1<br />
= k + 2<br />
Condition II also holds. Thus, formula holds true for all natural numbers.<br />
_________________________________________________________________________________________________<br />
15. The yearly values of <strong>the</strong> Durango form a<br />
geometric sequence with first term a 1 = 31,000<br />
and common ratio r = 0.85 (which represents a<br />
15% loss in value).<br />
1<br />
a 31,000 ( 0.85)<br />
n −<br />
n = ⋅<br />
The nth term of <strong>the</strong> sequence represents <strong>the</strong><br />
value of <strong>the</strong> Durango at <strong>the</strong> beginning of <strong>the</strong> nth<br />
year. Since we want to know <strong>the</strong> value after 10<br />
years, we are looking for <strong>the</strong> 11 th term of <strong>the</strong><br />
sequence. That is, <strong>the</strong> value of <strong>the</strong> Durango at <strong>the</strong><br />
beginning of <strong>the</strong> 11 th year.<br />
11 1<br />
a ( ) 10<br />
11 = a1 ⋅ r − = 31,000 ⋅ 0.85 = 6,103.11<br />
After 10 years, <strong>the</strong> Durango will be worth<br />
$6,103.11.<br />
16. The weights for each set form an arithmetic<br />
sequence with first term a 1 = 100 and common<br />
difference d = 30 . If we imagine <strong>the</strong> weightlifter<br />
only performed one repetition per set, <strong>the</strong> total<br />
weight lifted in 5 sets would be <strong>the</strong> sum of <strong>the</strong><br />
first five terms of <strong>the</strong> sequence.<br />
an<br />
= a1<br />
+ ( n−1)<br />
d<br />
a5<br />
= 100 + ( 5 − 1)( 30) = 100 + 4( 30)<br />
= 220<br />
n<br />
Sn<br />
= ( a+<br />
an)<br />
2<br />
S 5<br />
5 = ( 100 + 220) = 5<br />
2 2( 320)<br />
= 800<br />
Since he performs 10 repetitions in each set, we<br />
multiply <strong>the</strong> sum by 10 to obtain <strong>the</strong> total weight<br />
lifted.<br />
10( 800)<br />
= 8000<br />
The weightlifter will have lifted a total of 8000<br />
pounds after 5 sets.<br />
<strong>12</strong>83<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
<strong>Chapter</strong> <strong>12</strong> Cumulative Review<br />
1.<br />
2. a.<br />
2<br />
x = 9<br />
2 2<br />
x = 9 or x =−9<br />
x =± 3 or x =± 3i<br />
The solution set is { −3,3, − 3,3} i i .<br />
b.<br />
x<br />
2 2<br />
+ y = 100 and<br />
2<br />
y = 3x<br />
.<br />
2 2<br />
⎪<br />
⎧ x + y =<br />
3<br />
⎯⎯→<br />
2<br />
x +<br />
2<br />
y =<br />
2 2<br />
100 3 3 300<br />
⎨<br />
⎪⎩<br />
y = 3 x ⎯⎯→− 3 x + y = 0<br />
2<br />
3y<br />
+ y = 300<br />
2<br />
2<br />
3y<br />
+ y− 300=<br />
0<br />
( )( )<br />
( )<br />
− 1± 1 −4 3 −300 − 1±<br />
3601<br />
y = =<br />
23 6<br />
Substitute and solve for x:<br />
− 1+ 3601 2 − 1+<br />
3601<br />
y = ⇒ 3x<br />
=<br />
6 6<br />
2 − 1+ 3601 − 1+<br />
3601<br />
x = ⇒ x =±<br />
18 18<br />
or<br />
−1−<br />
3601 2 −1−<br />
3601<br />
y = ⇒ 3x<br />
=<br />
6 6<br />
2 −1−<br />
3601 −1−<br />
3601<br />
x = ⇒ x = ±<br />
18 18<br />
−1−<br />
3601<br />
which is not real since < 0<br />
18<br />
Therefore, <strong>the</strong> system has solutions:<br />
⎧⎛<br />
⎪ − 1+ 3601 − 1+<br />
3601<br />
⎞<br />
⎨ ⎜<br />
, ⎟ ,<br />
⎪ ⎜ 18 6 ⎟<br />
⎩ ⎝<br />
⎠<br />
⎛<br />
⎜−<br />
⎜<br />
⎝<br />
− 1+ 3601 − 1+<br />
3601 ⎞⎫<br />
, ⎟ ⎪ ⎬<br />
18 6 ⎟<br />
⎠⎭ ⎪<br />
c. The graphs of <strong>the</strong> circle and parabola<br />
intersect at <strong>the</strong> points<br />
⎛ − 1+ 3601 − 1+<br />
3601 ⎞<br />
⎜<br />
, ⎟≈ 1.81,9.84<br />
⎜ 18 6 ⎟<br />
⎝<br />
⎠<br />
⎛<br />
−<br />
⎜<br />
⎝<br />
x<br />
3. 2e = 5<br />
x 5<br />
e =<br />
2<br />
x ⎛5<br />
⎞<br />
ln ( e ) = ln ⎜ ⎟<br />
⎝ 2 ⎠<br />
⎛5<br />
⎞<br />
x = ln ⎜ ⎟≈0.916<br />
⎝2<br />
⎠<br />
− 1+ 3601 − 1+<br />
3601 ⎞<br />
, ⎟≈ −<br />
18 6 ⎟<br />
⎠<br />
⎧ ⎛5<br />
⎞⎫<br />
The solution set is ⎨ln ⎜ ⎟⎬<br />
⎩ ⎝ 2 ⎠⎭ .<br />
( )<br />
( 1.81,9.84)<br />
4. slope = m = 5 ; Since <strong>the</strong> x-intercept is 2, we<br />
know <strong>the</strong> point ( 2,0 ) is on <strong>the</strong> graph of <strong>the</strong> line<br />
and is a solution to <strong>the</strong> equation y = 5x+ b.<br />
y = 5x+<br />
b<br />
0= 5( 2)<br />
+ b<br />
0= 10+<br />
b<br />
− 10 = b<br />
Therefore, <strong>the</strong> equation of <strong>the</strong> line with slope 5<br />
and x-intercept 2 is y = 5x− 10.<br />
5. Given a circle with center (–1, 2) and containing<br />
<strong>the</strong> point (3, 5), we first use <strong>the</strong> distance formula<br />
to determine <strong>the</strong> radius.<br />
( ( )) ( )<br />
2 2 2 2<br />
r = 3− − 1 + 5− 2 = 4 + 3<br />
= 16 + 9 = 25 = 5<br />
Therefore, <strong>the</strong> equation of <strong>the</strong> circle is given by<br />
2 2 2<br />
x− − 1 + y− 2 = 5<br />
( ( )) ( )<br />
( x ) ( y )<br />
2 2 2<br />
+ 1 + − 2 = 5<br />
2 2<br />
x + 2x+ 1+ y − 4y+ 4=<br />
25<br />
2 2<br />
x + y + 2x−4y− 20=<br />
0<br />
<strong>12</strong>84<br />
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<strong>Chapter</strong> <strong>12</strong> Cumulative Review<br />
3x<br />
x 2<br />
6. f ( x)<br />
= , g ( x) = 2x+<br />
1<br />
−<br />
a. g ( ) ( )<br />
f ( )<br />
( )<br />
2 = 2 2 + 1=<br />
5<br />
35 15<br />
5 = = = 5<br />
5−<br />
2 3<br />
( f g)( ) f ( g( )) f ( )<br />
b. f ( )<br />
2 = 2 = 5 = 5<br />
( )<br />
34 <strong>12</strong><br />
4 = = = 6<br />
4−<br />
2 2<br />
g ( 6) = 2( 6)<br />
+ 1=<br />
13<br />
( g<br />
f )( ) g( f ( )) g( )<br />
4 = 4 = 6 = 13<br />
c. ( f g)( x) = f ( g( x)<br />
)<br />
32 ( x + 1)<br />
=<br />
( x + ) −<br />
2 1 2<br />
6x<br />
+ 3<br />
=<br />
2x<br />
−1<br />
d. To determine <strong>the</strong> domain of <strong>the</strong> composition<br />
( f g)( x)<br />
, we start with <strong>the</strong> domain of g<br />
and exclude any values in <strong>the</strong> domain of g<br />
that make <strong>the</strong> composition undefined.<br />
g x is defined for all real numbers and<br />
( )<br />
( f g)( x)<br />
is defined for all real numbers<br />
except<br />
1<br />
x = . Therefore, <strong>the</strong> domain of <strong>the</strong><br />
2<br />
⎧ 1 ⎫<br />
composite ( f g)( x)<br />
is<br />
e. ( g<br />
f )( x)<br />
⎛ 3x<br />
⎞<br />
= 2⎜<br />
⎟+<br />
1<br />
⎝ x − 2 ⎠<br />
6x<br />
= + 1<br />
x − 2<br />
6x+ x−2<br />
=<br />
x − 2<br />
7x<br />
− 2<br />
=<br />
x − 2<br />
⎨xx≠<br />
⎬<br />
⎩ 2⎭ .<br />
f. To determine <strong>the</strong> domain of <strong>the</strong> composition<br />
( g f )( x)<br />
, we start with <strong>the</strong> domain of f<br />
and exclude any values in <strong>the</strong> domain of f<br />
that make <strong>the</strong> composition undefined.<br />
f ( x ) is defined for all real numbers except<br />
2 g f x is defined for all real<br />
x = and ( )( )<br />
numbers except x = 2 . Therefore, <strong>the</strong><br />
domain of <strong>the</strong> composite ( g f )( x)<br />
is<br />
{ x| x ≠ 2}<br />
.<br />
g. g ( x) = 2x+<br />
1<br />
1<br />
2<br />
y = 2x+<br />
1<br />
x = 2y<br />
+ 1<br />
x − 1=<br />
2y<br />
( x 1)<br />
− = y<br />
−<br />
g ( x) = ( x−<br />
1)<br />
2<br />
−1<br />
The domain of g ( x)<br />
1 1<br />
numbers.<br />
3x<br />
f x =<br />
x − 2<br />
3x<br />
y =<br />
x − 2<br />
3y<br />
x =<br />
y − 2<br />
x( y− 2)<br />
= 3y<br />
xy− 2x = 3y<br />
xy− 3y = 2x<br />
y( x− 3)<br />
= 2x<br />
2x<br />
y =<br />
x − 3<br />
−1 2x<br />
f ( x)<br />
=<br />
x − 3<br />
f<br />
h. ( )<br />
−1<br />
The domain of ( x)<br />
is <strong>the</strong> set of all real<br />
is { x| x ≠ 3}<br />
.<br />
7. Center: (0, 0); Focus: (0, 3); Vertex: (0, 4);<br />
Major axis is <strong>the</strong> y-axis; a = 4; c = 3 .<br />
2 2 2<br />
Find b: b = a − c = 16 − 9 = 7 ⇒ b=<br />
7<br />
Write <strong>the</strong> equation using rectangular coordinates:<br />
2 2<br />
x y<br />
+ = 1<br />
7 16<br />
Parametric equations for <strong>the</strong> ellipse are:<br />
x = 7cos π t ; y = 4sin π t ; 0≤t<br />
≤ 2<br />
( ) ( )<br />
<strong>12</strong>85<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
8. The focus is ( − 1, 3)<br />
and <strong>the</strong> vertex is ( 1, 2)<br />
− .<br />
Both lie on <strong>the</strong> vertical line x =− 1. We have<br />
a = 1 since <strong>the</strong> distance from <strong>the</strong> vertex to <strong>the</strong><br />
−1, 3 is above<br />
focus is 1 unit, and since ( )<br />
( − 1, 2)<br />
, <strong>the</strong> parabola opens up. The equation of<br />
<strong>the</strong> parabola is:<br />
2<br />
( x− h) = 4a( y−k)<br />
2<br />
( x−( − 1)<br />
) = 4⋅1⋅( y−2)<br />
2<br />
( x+ 1) = 4( y−2)<br />
9. Center point (0, 4); passing through <strong>the</strong> pole<br />
(0,4) implies that <strong>the</strong> radius = 4 using rectangular<br />
coordinates:<br />
2 2 2<br />
( x − h) + ( y− k)<br />
= r<br />
2 2 2<br />
x− 0 + y− 4 = 4<br />
10.<br />
( ) ( )<br />
2<br />
x ( y )<br />
+ − 4 = 16<br />
2 2<br />
x + y − 8y+ 16=<br />
16<br />
2 2<br />
x + y − 8y<br />
= 0<br />
converting to polar coordinates:<br />
2<br />
r − 8rsinθ<br />
= 0<br />
2<br />
r = 8rsinθ<br />
r = 8sinθ<br />
2<br />
2<br />
2sin x−sin x− 3= 0, 0≤ x≤<br />
2π<br />
( x )( x )<br />
2sin − 3 sin + 1 = 0<br />
3<br />
2sin x− 3 = 0 ⇒ sin x = , which is impossible<br />
2<br />
3π<br />
sin x+ 1 = 0 ⇒ sin x = −1⇒ x =<br />
2<br />
⎧3π<br />
⎫<br />
The solution set ⎨ ⎬<br />
⎩ 2 ⎭ .<br />
−1<br />
11. cos ( − 0.5)<br />
We are finding <strong>the</strong> angle θ, −π ≤θ ≤ π,<br />
whose<br />
cosine equals − 0.5 .<br />
cosθ =−0.5<br />
−π ≤θ ≤π<br />
2π<br />
− 1 2π<br />
θ = ⇒cos ( − 0.5)<br />
=<br />
3 3<br />
<strong>12</strong>.<br />
1<br />
sin θ = , θ is in Quadrant II<br />
4<br />
a. θ is in Quadrant II ⇒ cosθ<br />
< 0<br />
b.<br />
⎞<br />
cosθ<br />
=− 1− sin θ =− 1− ⎜ ⎟<br />
⎝4<br />
⎠<br />
1 15 15<br />
=− 1− =− =−<br />
16 16 4<br />
2 ⎛1<br />
⎛1<br />
⎞<br />
sinθ<br />
⎜ ⎟<br />
4 ⎛1 ⎞⎛ 4 ⎞<br />
tanθ<br />
= =<br />
⎝ ⎠<br />
= ⎜ ⎟⎜ − ⎟<br />
cosθ<br />
⎛ 15 ⎞ ⎝4 ⎠⎝ 15 ⎠<br />
⎜<br />
−<br />
4 ⎟<br />
⎝ ⎠<br />
1 15 15<br />
=− ⋅ =−<br />
15 15 15<br />
c. sin(2 θ ) = 2sinθcosθ<br />
⎛1⎞⎛ 15⎞<br />
= 2⎜ ⎜−<br />
⎝ 4<br />
⎟⎜ 4 ⎟<br />
⎠⎝ ⎠<br />
15<br />
=−<br />
8<br />
d.<br />
e.<br />
2 2<br />
cos(2 θ) = cos θ −sin<br />
θ<br />
2 2<br />
⎛ 15 ⎞ ⎛1<br />
⎞<br />
= ⎜<br />
− 4 ⎟<br />
−⎜ ⎟<br />
⎝ ⎠ ⎝4⎠<br />
15 1 14 7<br />
= − = =<br />
16 16 16 8<br />
π π θ π<br />
< θ < π ⇒ < <<br />
2 4 2 2<br />
θ<br />
⎛θ<br />
⎞<br />
⇒ is in Quadrant I ⇒ sin ⎜ ⎟><br />
0<br />
2<br />
⎝2<br />
⎠<br />
⎛ 15 ⎞<br />
1− θ 1 cosθ<br />
⎜<br />
−<br />
4 ⎟<br />
⎛ ⎞ −<br />
sin<br />
⎝ ⎠<br />
⎜ ⎟= =<br />
⎝ 2 ⎠ 2 2<br />
⎛4+<br />
15⎞<br />
⎜<br />
4 ⎟<br />
4+<br />
15<br />
=<br />
⎝ ⎠<br />
=<br />
2 8<br />
4+<br />
15<br />
=<br />
2 2<br />
2<br />
<strong>12</strong>86<br />
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<strong>Chapter</strong> <strong>12</strong> Projects<br />
<strong>Chapter</strong> <strong>12</strong> Projects<br />
Project I<br />
Answers will vary based on <strong>the</strong> year that is used. Data<br />
used in <strong>the</strong>se solutions will be from 2001.<br />
1. Race Birth rate Death rate<br />
White 0.0118 0.00991<br />
African-American 0.116 0.007758<br />
American Indian/<br />
Alaska native<br />
0.0137 0.00392<br />
Asian/Pacific<br />
Islander<br />
0.0164 0.003047<br />
Hispanic 0.0230 0.00306<br />
Total 0.141 0.008485<br />
I = net immigration = 973,206<br />
Population for 2001 = 285,545,000<br />
2. r = 0.0141 – 0.008485 = 0.005615<br />
3. pn<br />
= (1 + 0.005615) pn−1<br />
+ 973206<br />
pn<br />
= (1.005615) pn−1<br />
+ 973206<br />
p = 285545000<br />
0<br />
4. p1 p0<br />
= (1.005615) + 973206<br />
p1<br />
= (1.005615)(285545000) + 973206<br />
p = 288,<strong>12</strong>1,541<br />
1<br />
The population is predicted to be 288,<strong>12</strong>1,541 in<br />
2002.<br />
5. Actual population in 2002: 288,600,000. The<br />
formula’s prediction was lower but fairly close.<br />
6. Birth rate: 47.52 per 1000 population (0.04752)<br />
Death rate: 17.97 per 1000 population (0.01797)<br />
Population for 2001: 23,985,7<strong>12</strong><br />
I = net immigration = − 6956<br />
r = 0.04752 − 0.01797 = 0.02955<br />
pn<br />
= (1 + 0.02955) pn−1<br />
−6956<br />
pn<br />
= (1.02955) pn−1<br />
−6956<br />
p0<br />
= 23,985,7<strong>12</strong><br />
p1 = (1.02955) p0<br />
−6956<br />
p1<br />
= (1.02955)(239857<strong>12</strong>) −6956<br />
p1<br />
= 24,687,534<br />
The population is predicted to be 24,687,534 in<br />
2002.<br />
Actual population in 2002: 24,699,073.<br />
The formula’s prediction was higher but fairly<br />
close.<br />
7. Answers will vary. This appears to support <strong>the</strong><br />
article. The growth rate for <strong>the</strong> U.S. is much<br />
smaller than <strong>the</strong> growth rate for Uganda.<br />
8. It could be but one must consider trends in each<br />
of <strong>the</strong> pieces of data to find if <strong>the</strong> growth rate is<br />
increasing or decreasing over time. The same<br />
thing must be examined with respect to <strong>the</strong> net<br />
immigration.<br />
Project II<br />
1. 2, 4, 8, 16, 32, 64<br />
2. length n 2 n levels<br />
This is a geometric sequence: a 2 n<br />
n =<br />
Recursive expression: an<br />
= 2 an−1, a0<br />
= 1<br />
3.<br />
n<br />
256 = 2<br />
8 n<br />
2 = 2<br />
n = 8<br />
Project III<br />
1. Qst = − 3+ 2 Pt−<br />
1, Qdt = 18−3Pt<br />
P0<br />
= 2, b= 2, d = 3, c = 18<br />
− a = −3 → a = 3<br />
2.<br />
3+ 18−2P<br />
21−2P<br />
Pt<br />
= =<br />
3 3<br />
2<br />
Pt<br />
= 7 − Pt−<br />
1, P0<br />
= 2<br />
3<br />
−10<br />
t−1 t−1<br />
7<br />
−9<br />
P t<br />
10<br />
Pt<br />
− 1<br />
<strong>12</strong>87<br />
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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />
3. P1<br />
=<br />
Q<br />
Q<br />
17<br />
3<br />
=− 3+ 2(2) Q<br />
⎛17<br />
⎞<br />
= 18−3 ⎜ ⎟<br />
⎝ 3 ⎠<br />
= 1 Q = 1<br />
s1 d1<br />
s1 d1<br />
29<br />
P2<br />
=<br />
9<br />
⎛17 ⎞ ⎛29<br />
⎞<br />
Qs2 =− 3+ 2⎜ ⎟ Qd2<br />
= 18−3⎜ ⎟<br />
⎝ 3 ⎠ ⎝ 9 ⎠<br />
25 25<br />
Qs2 = Qd2<br />
=<br />
3 3<br />
The market (supply and demand) are getting<br />
closer to being <strong>the</strong> same.<br />
4. The equilibrium price is 4.20.<br />
5. It takes 17 time periods.<br />
6. Qd17<br />
Q<br />
s17<br />
Project IV<br />
= 18 − 3(4.20) = 5.40<br />
=− 3+ 2(4.20) = 5.40<br />
The equilibrium quantity is 5.4.<br />
1. 1, 2, 4, 7, 11, 16, 22, 29<br />
2. It is not arithmetic because <strong>the</strong>re is no common<br />
difference. It is not geometric because <strong>the</strong>re is no<br />
common ration.<br />
3. 15<br />
−2<br />
−2<br />
4. y = 2.5x−<br />
2.5<br />
The graph does not pass through any of <strong>the</strong><br />
points.<br />
y6<br />
= <strong>12</strong>.5<br />
y7<br />
= 15<br />
y = 17.5<br />
8<br />
6<br />
5.<br />
y1<br />
= 0<br />
y2<br />
= 2.5<br />
y3<br />
= 5<br />
y4<br />
= 7.5<br />
y = 10<br />
5<br />
5<br />
∑<br />
i=<br />
1<br />
( y − y )<br />
r<br />
i<br />
i<br />
= (0 − 1) + (2.5 − 2) + (5 − 4) + (7.5 − 7) + (10 −11)<br />
=−1<br />
This is <strong>the</strong> sum of <strong>the</strong> errors.<br />
2<br />
y = 0.5x − 0.5x+<br />
1<br />
The graph passes through all of <strong>the</strong> points.<br />
y6<br />
= 16<br />
y7<br />
= 22<br />
y = 29<br />
8<br />
y1<br />
= 1<br />
y2<br />
= 2<br />
y3<br />
= 4<br />
y4<br />
= 7<br />
y = 11<br />
5<br />
5<br />
∑<br />
i=<br />
1<br />
( y − y ) = 0<br />
r<br />
i<br />
i<br />
The sum of <strong>the</strong> errors is zero.<br />
6. When trying to obtain <strong>the</strong> cubic and quartic<br />
polynomials of best fit, <strong>the</strong> cubic and quartic<br />
terms have coefficient zero and <strong>the</strong> polynomial<br />
of best fit is given as <strong>the</strong> quadratic in part e.<br />
For <strong>the</strong> exponential function of best fit,<br />
(0.59)(1.83) x<br />
y = .<br />
y6 = 22.2 y7 = 40.6 y8<br />
= 74.2<br />
The sum of <strong>the</strong>se errors becomes quite large.<br />
This error shows that <strong>the</strong> function does not fit <strong>the</strong><br />
data very well as x gets larger.<br />
7. The quadratic function is best.<br />
8. The data does not appear to be ei<strong>the</strong>r logarithmic<br />
or sinusoidal in shape, so it does not make sense<br />
to try to fit one of those functions to <strong>the</strong> data.<br />
<strong>12</strong>88<br />
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