Chapter 12 Sequences; Induction; the Binomial Theorem

Chapter 12 

Sequences; Induction; the Binomial Theorem 

Section 12.1 

2−1 1 

2 2 

1. f ( 2) 

= = ; f ( 3) 

2. True 

3. 

4. 

nt ⋅ 

3−1 2 

= = 

3 3 

⎛ r ⎞ 

A= P⎜1+ 

⎟ 

⎝ n ⎠ 

22 ⋅ 

⎛ 0.04 ⎞ 

= 1000⎜1+ 

⎟ 

⎝ 2 ⎠ 

4 

= 1000( 1.02) 

= 1082.43 

After two years, the account will contain 

$1082.43. 

nt ⋅ 

⎛ r ⎞ 

A= P⎜1+ 

⎟ 

⎝ n ⎠ 

⎛ 0.05 ⎞ 

10,000 = P ⎜1+ 

⎟ 

⎝ 12 ⎠ 

12⋅1 

⎛ 0.05 ⎞ 

10,000 = P⎜1+ 

⎟ 


10,000 = P( 1.051162) 

10,000 

= P 

1.051162 

9513.28 = P 

To have $10,000 at the end of one year, you need 

to invest $9513.28 now. 

5. sequence 

12 

6. s 1 = 41 () − 1= 3; ( ) 

4 4 

7. ∑( k) 

∑ 

k= 1 k= 

1 

8. True 

s 4 = 4 4 − 1= 

15 

( + ) 

4 4 1 

2 = 2 k = 2⋅ = 4( 5) 

= 20 

2 

9. True; a sequence is a function whose domain is 

the set of positive integers. 

11. 10! = 10987654321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 3,628,800 

12. 9! = 987654321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 362,880 

13. 

14. 

15. 

16. 

9! 9⋅8⋅7⋅6! 

= = 987 ⋅ ⋅ = 504 

6! 6! 

12! 12⋅11⋅10! 

= = 12⋅ 11 = 132 

10! 10! 

3!7! ⋅ 3217654! 

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 

= = 321765 ⋅ ⋅ ⋅ ⋅ ⋅ = 1260 

4! 4! 

5! ⋅8! 5⋅4⋅3! ⋅8! 

= 

3! 3! 

= 5487654321 

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 

= 806,400 

17. s1 s2 s3 s4 s5 

18. 

= 1, = 2, = 3, = 4, = 5 

2 2 2 

1 2 3 

2 2 

4 = 4 + 1= 17, s5 

= 5 + 1= 

26 

s = 1 + 1= 2, s = 2 + 1= 5, s = 3 + 1= 

10, 

s 

1 1 2 2 1 

19. a1 = = , a2 

= = = , 

1+ 2 3 2+ 

2 4 2 

3 3 4 4 2 

a3 = = , a4 

= = = , 

3+ 2 5 4+ 

2 6 3 

5 5 

a 5 = = 

5+ 

2 7 

21 ⋅ + 1 3 22 ⋅ + 1 5 

20. b1 = = , b2 

= = , 

21 ⋅ 2 22 ⋅ 4 

23 ⋅ + 1 7 24 ⋅ + 1 9 

b3 = = , b4 

= = , 

23 ⋅ 6 24 ⋅ 8 

25 ⋅ + 1 11 

b5 

= = 

25 ⋅ 10 

21. 

11 + 2 21 + 2 

1 c2 

3+ 1 2 4+ 

1 2 

3 c4 

5+ 

1 2 

c = ( − 1) (1 ) = 1, = ( − 1) (2 ) =−4, 

c = ( − 1) (3 ) = 9, = ( − 1) (4 ) = −16, 

c 5 = ( − 1) (5 ) = 25 

10. True; 

2 

∑ 

k = 1 

( + ) 

2 2 1 

k = = 3 

2 

1235 

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Chapter 12: Sequences; Induction; the Binomial Theorem 

22. 

23. 

24. 

25. 

26. 

11 − ⎛ 1 ⎞ 

d1 

= ( − 1) ⎜ ⎟= 

1, 

⎝21 ⋅ −1⎠ 

2−1⎛ 

2 ⎞ 2 

d2 

= ( − 1) ⎜ ⎟= − , 

⎝2⋅2−1⎠ 

3 

3−1⎛ 

3 ⎞ 3 

d3 

= ( − 1) ⎜ ⎟= 

, 

⎝23 ⋅ −1⎠ 

5 

d 

d 

4 

5 

4−1⎛ 

4 ⎞ 4 

= ( − 1) ⎜ ⎟= − , 

⎝2⋅4−1⎠ 

7 

5−1⎛ 

5 ⎞ 5 

= ( − 1) ⎜ ⎟= 

⎝25 ⋅ −1⎠ 

9 

1 2 

2 2 1 2 4 2 

s1 = = = , s 

1 2 = = = , 

2 

3 + 1 4 2 3 + 1 10 5 

s 

3 4 

2 8 2 2 16 8 

= = = , s = = = , 

3 + 1 28 7 3 + 1 82 41 

3 3 4 4 

s 

5 5 

5 

2 32 8 

= = = 

3 + 1 244 61 

1 2 

⎛4⎞ 4 ⎛4⎞ 

16 

s1 = ⎜ ⎟ = , s2 

= ⎜ ⎟ = , 

⎝3⎠ 3 ⎝3⎠ 

9 

s 

s 

3 4 

3 4 

5 

⎛4 ⎞ 64 ⎛4 ⎞ 256 

= ⎜ ⎟ = , s = ⎜ ⎟ = , 

⎝3⎠ 27 ⎝3⎠ 

81 

5 

⎛4⎞ 

1024 

= ⎜ ⎟ = 

⎝3⎠ 

243 

1 

( −1) −1 1 

t1 

= = = − , 

(1+ 1)(1 + 2) 2 ⋅3 6 

t 

t 

t 

t 

2 

3 

4 

5 

2 

( −1) 1 1 

= = = , 

(2 + 1)(2 + 2) 3⋅4 12 

3 

( −1) −1 1 

= = = − , 

(3 + 1)(3 + 2) 4⋅5 20 

4 

( −1) 1 1 

= = = , 

(4 + 1)(4 + 2) 5⋅6 30 

5 

( −1) −1 1 

= = =− 

(5 + 1)(5 + 2) 6⋅7 42 

1 2 3 

3 3 3 9 3 27 

a1 = = = 3, a2 = = , a3 

= = = 9, 

1 1 2 2 3 3 

a 

4 5 

3 81 3 243 

= = , a = = 

4 4 5 5 

4 5 

1 1 2 3 4 5 

e e e e e e 

27. b1 = = , b 

1 2 = , b 

2 3 = , b 

3 4 = , b 

4 5 = 5 

28. 

2 2 2 

1 1 2 2 3 3 

1 1 2 3 9 

c = = , c = = 1, c = = , 

2 2 2 2 8 

2 2 

4 16 5 25 

c4 = = = 1, c 

4 5 = = 

5 

2 16 2 32 

29. Answers may vary. One possibility follows: 

Each term is a fraction with the numerator equal 

to the term number and the denominator equal to 

one more than the term number. 

n 

an 

= 

n + 1 



to 1 and the denominator equal to the product of 

the term number and one more than the term 

number. 

1 

a n = 

n n+ 

1 

( ) 



to 1 and the denominator equal to a power of 2. 

The power is equal to one less than the term 

number. 

1 

a n = 

n−1 

2 


Each term is equal to a fraction with the 

numerator equal to a power of 2 and the 

denominator equal to a power of 3. Both powers 

are equal to the term number. Since the powers 

are the same, we can use rules for exponents to 

write each term as a power of 2 3 . 

a n 

⎛2 

⎞ 

= ⎜ ⎟ 

⎝3 

⎠ 

n 


The terms form an alternating sequence. Ignoring 

the sign, each term always contains a 1. The sign 

alternates by raising − 1 to a power. Since the 

first term is positive, we use n − 1 as the power. 

1 

1 n − 

= − 

a n 

( ) 




Section 12.1: Sequences 


The terms appear to alternate between whole 

numbers and fractions. If we write the whole 

numbers as fractions (e.g. 1 = 1 , 3 = 3 , etc.), we 

1 1 

see that each term consists of a 1 and the term 

number. When n is odd, the numerator is n and the 

denominator is 1. When n is even, the numerator is 

1 and the denominator is n. This alternating 

behavior occurs if we have a power that alternates 

sign. The alternating sign is obtained by using 

( − ) 

1 n+1 

. Thus, we get 

an 

( − 

= n 

) 

1 

1 n + 


The terms (ignoring the sign) are equal to the 

term number. The alternating sign is obtained by 

using ( − 1) 

n+1 

. 

an 

( ) 

1 

1 n + 

= − ⋅ n 


Here again we have alternating signs which will 

be taken care of by using ( − 1) 

n+1 

. The rest of the 

term is twice the term number. 

an 

( ) 

n+ 

1 

= −1 ⋅ 2n 

37. a1 a2 a3 

= 2, = 3 + 2 = 5, = 3 + 5 = 8, 

a = 3 + 8 = 11, a = 3+ 11 = 14 

4 5 

a = 3, a = 4 − 3 = 1, a = 4 − 1 = 3, 

a = 4− 3= 1, a = 4− 1= 

3 

38. 1 2 3 

4 5 

39. a1 a2 a3 

=− 2, = 2 + ( − 2) = 0, = 3 + 0 = 3, 

a = 4+ 3= 7, a = 5+ 7 = 12 

4 5 

40. a1 a2 a3 

= 1, = 2 − 1 = 1, = 3 − 1 = 2, 

a = 4− 2= 2, a = 5− 2= 

3 

4 5 

a = 5, a = 2⋅ 5 = 10, a = 2⋅ 10 = 20, 

a = 220 ⋅ = 40, a = 240 ⋅ = 80 

41. 1 2 3 

4 5 

42. a1 a2 a3 

= 2, = − 2, = −( − 2) = 2, 

a =− 2, a =−− ( 2) = 2 

4 5 

3 1 1 

3 

43. 2 1 2 1 8 1 

a1= 3, a2 = , a3 = = , a4 

= = , a5 

= = 

2 3 2 4 8 5 40 

44. a1 = − 2, a2 

= 2 + 3( − 2) =−4, 

a3 = 3 + 3( − 4) = − 9, a4 

= 4 + 3( − 9) =−23, 

a 5 = 5+ 3( − 23) =− 64 

= 1, = 2, = 2⋅ 1 = 2, = 2⋅ 2 = 4, 

a = 42 ⋅ = 8 

45. a1 a2 a3 a4 

5 

46. a1 a2 a3 

= − 1, = 1, =− 1+ 3⋅ 1 = 2, 

a = 1+ 4⋅ 2= 9, a = 2+ 5⋅ 9= 

47 

4 5 

a = A, a = A+ d, a = ( A+ d) + d = A+ 

2 d, 

a4 

= ( A+ 2 d) + d = A+ 

3 d, 

a = ( A+ 3 d) + d = A+ 

4d 

47. 1 2 3 

48. 

5 

2 

1 = , 2 = , 3 = ( ) = , 

2 3 3 4 

4 = = 5 = = 

a A a rA a r rA r A 

( ) , ( ) 

a r r A r A a r r A r A 

49. a1 a2 a3 

a 

4 

= 2, = 2+ 2, = 2+ 2+ 

2, 

= 2+ 2+ 2+ 

2 , 

a 5 = 2+ 2+ 2+ 2+ 

2 

2 

2 

= 2, = , = 

2 

, 

2 2 

50. a1 a2 a3 

n 

a 

2 

2 2 

2 2 

= 

2 

, a = 

2 

2 2 

4 5 

51. ∑ ( k+ 2) = 3+ 4+ 5+ 6+ 7+⋅⋅⋅+ ( n+ 

2) 

k = 1 

n 

52. ∑ (2k 

+ 1) = 3+ 5 + 7 + 9 +⋅⋅⋅+ ( 2n+ 

1) 

53. 

k = 1 

n 

∑ 

k = 1 

n 

2 2 

k 1 9 25 49 n 

= + 2 + + 8 + + 18 + + 32 +⋅⋅⋅+ 

2 2 2 2 2 2 

2 2 

54. ∑ ( k+ 1) = 4+ 9+ 16+ 25+ 36+⋅⋅⋅+ ( n+ 

1) 

k = 1 





55. 

56. 

57. 

n 

1 1 1 1 1 

∑ = 1+ + + + + 

3 

k 3 9 27 3 

n 

k = 0 

n 

⎛3⎞ 3 9 ⎛3⎞ 

∑ ⎜ ⎟ = 1+ + + + ⎜ ⎟ 

⎝2⎠ 2 4 ⎝2⎠ 

k = 0 

n−1 

k = 0 

k 

1 1 1 1 1 

∑ = + + + + 

k+ 

1 

3 3 9 27 

n 

3 

n−1 

58. ∑ (2k 

+ 1) = 1+ 3+ 5+ 7 + + ( 2( n− 1) + 1) 

59. 

60. 

k = 0 

n 

k = 2 

= 1+ 3+ 5+ 7 + + (2n 

−1) 

k 

n 

∑ ( − 1) lnk 

= ln2− ln3+ ln4 − + ( −1) lnn 

n 

∑ 

k = 3 

k+ 

1 k 

( −1) 2 

4 3 5 4 6 5 n+ 

1 n 

= ( − 1) 2 + ( − 1) 2 + ( − 1) 2 + + ( −1) 2 

3 4 5 6 n+ 

1 n 

= 2 − 2 + 2 − 2 + + ( −1) 2 

n+ 

1 n 

= 8− 16+ 32− 64 + ... + ( −1) 2 


20 

1+ 2+ 3+ + 20= ∑ k 

k = 1 


8 

3 3 3 3 3 

1 + 2 + 3 + + 8 = ∑ k 

k = 1 


1 2 3 13 

13 

k 

+ + + + = ∑ 

2 3 4 13+ 1 k + 1 

k = 1 



1+ 3 + 5 + 7 + + 2(12) − 1 = (2k 

−1) 

[ ] 

n 

∑ 

k = 1 


1 1 1 

6 

6 ⎛ 1 ⎞ k ⎛ 1 ⎞ 

1 − + − + + ( − 1) ⎜ ( 1) 

3 9 27 

6 ⎟= ∑ − ⎜ k ⎟ 

⎝3 ⎠ ⎝3 

⎠ 

k = 0 


11 

2 4 8 

11 

11+ 1 ⎛2⎞ k + 1 ⎛2⎞ 

− + + + ( − 1) ⎜ ⎟ = ∑ ( −1) 

⎜ ⎟ 

3 9 27 

⎝3⎠ ⎝3⎠ 

k = 1 

k 


2 3 

n k 

3 3 3 3 

3 + n 

2 + 3 

+ + n 

= ∑ k 

k = 1 


1 2 3 n 

n 

k 

+ + + + = 

e 

2 3 

n 

∑ 

k 

e e e e 

k = 1 


n 

a+ ( a+ d) + ( a+ 2 d) + + ( a+ nd) = ( a+ 

kd) 

n 

∑ 

k = 1 

( ) 

or = ( a+ k −1 d) 

∑ 

k = 0 


n 

2 n−1 k−1 

a+ ar+ ar + + ar = ∑ ar 

or 

n−1 

∑ 

k = 0 

ar 

k 

k = 1 

40 

71. ( ) 

72. 

73. 

74. 

75. 

∑ 5= 5 

+ 5+ 5+⋅⋅⋅+ 5= 40 5 = 200 

k = 1 40 times 

50 

∑ 8 = 8 

+ 8 + 8 +⋅⋅⋅+ 8 = 50(8) = 400 

k = 1 50 times 

40 

∑ 

k = 1 

( + ) 

40 40 1 

k = = 20( 41) 

= 820 

2 

24 24 

k= 1 k= 

1 

( + ) 

24 24 1 

( − k) = − k = − = −300 

2 

∑ ∑ 

20 20 20 20 20 

∑ ∑ ∑ ∑ ∑ 

(5k+ 3) = (5 k) + 3 = 5 k+ 

3 

k= 1 k= 1 k= 1 k= 1 k= 

1 

( + ) 

⎛20 20 1 ⎞ 

= 5⎜ 

⎟+ 

3 20 

⎝ 2 ⎠ 

= 1050 + 60 = 1110 

( ) 

26 26 26 26 26 

76. ∑ ( k − ) = ∑( k) 

− ∑ = ∑k−∑ 

3 7 3 7 3 7 

k= 1 k= 1 k= 1 k= 1 k= 

1 

( + ) 

⎛26 26 1 ⎞ 

= 3⎜ 

⎟−7 26 

⎝ 2 ⎠ 

= 1053 − 182 = 871 

( ) 





16 16 16 

2 2 

∑ ∑ ∑ 

77. ( k ) 

+ 4 = k + 4 

k= 1 k= 1 k= 

1 

( + )( ⋅ + ) 

16 16 1 2 16 1 

= + 416 

6 

= 1496 + 64 = 1560 

( ) 

14 14 

78. ∑ ( k 

2 − 4) = ( 0 2 − 4) + ∑( k 

2 −4) 

79. 

k= 0 k= 

1 

14 14 

2 

∑ 

= k − 

∑ 

k= 1 k= 

1 

4 

( + )( ⋅ + ) 

14 14 1 2 14 1 

=− 4+ −4 14 

6 

=− 4 + 1015− 64 = 955 

⎡ ⎤ 

2k = 2 2k = 2⎢ 

k − k⎥ 

⎣ ⎦ 

60 60 60 9 

∑ ∑ ∑ ∑ 

k= 10 k= 10 k= 1 k= 

1 

( + ) ( + ) 

⎡60 60 1 9 9 1 ⎤ 

= 2 ⎢ − ⎥ 

⎣ 2 2 ⎦ 

= 2 1830 − 45 = 3570 

[ ] 

⎡ ⎤ 

k k ⎢ k k⎥ 

⎣ ⎦ 

40 40 40 7 

80. ∑( − 3 ) =− 3∑ =−3 

∑ −∑ 

k= 8 k= 8 k= 1 k= 

1 

( + ) ( + ) 

⎡40 40 1 7 7 1 ⎤ 

=−3⎢ 

− ⎥ 

⎣ 2 2 ⎦ 

=−3 820 − 28 =−2376 

[ ] 

( ) 

From the table we see that the balance is 

below $2000 after 14 payments have been 

made. The balance then is $1953.70. 

c. Scrolling down the table, we find that 

balance is paid off in the 36th month. The 

last payment is $83.78. There are 35 

payments of $100 and the last payment of 

$83.78. The total amount paid is: 35(100) + 

83.78(1.01) = $3584.62. (we have to add 

the interest for the last month). 

d. The interest expense is: 

3584.62 – 3000.00 = $584.62 

81. 

82. 

20 20 4 

3 3 3 

∑ ∑ ∑ 

k = k − k 

k= 5 k= 1 k= 

1 

( + ) ( + ) 

2 2 

2 2 

⎡20 20 1 ⎤ ⎡4 4 1 ⎤ 

= ⎢ ⎥ −⎢ ⎥ 

⎣ 2 ⎦ ⎣ 2 ⎦ 

= 210 − 10 = 44,000 

24 24 3 

3 3 3 

∑ ∑ ∑ 

k = k − k 

k= 4 k= 1 k= 

1 

( + ) ( + ) 

2 2 

2 2 

⎡24 24 1 ⎤ ⎡3 3 1 ⎤ 

= ⎢ ⎥ −⎢ ⎥ 

⎣ 2 ⎦ ⎣ 2 ⎦ 

= 300 − 6 = 89,964 

84. a. B 1 = 1.005(18500) − 534.47 = $18,058.03 

b. Put the graphing utility in SEQuence mode. 

Enter Y= as follows, then examine the 

TABLE: 

83. a. B 1 = 1.01(3000) − 100 = $2930 

b. Put the graphing utility in SEQuence mode. 

Enter Y= as follows, then examine the 

TABLE: 

From the table we see that the balance is 

below $10,000 after 19 payments have been 

made. The balance then is $9713.76. 





c. Scrolling down the table, we find that 

balance is paid off in the 39th month. The 

last payment is $54.18. There are 38 

payments of $534.47 and the last payment 

of $54.18 plus interest. The total amount 

paid is: 38(534.47) + 54.18(1.005) = 

$20,364.31. 

d. The interest expense is: 

20,364.31 – 18,500.00 = $1864.31 

85. a. p1 

= 1.03(2000) + 20 = 2080; 

p2 

= 1.03(2080) + 20 = 2162.4 

There are approximately 2162 trout in the 

pond at the end of the second month. 

b. Scrolling down the table, we find the trout 

population exceeds 5000 at the end of the 

26th month when the population is 5084. 

c. The equilibrium level of pollution occurs 

when x = 0.9x+ 15 . That is, when x = 150 

tons. 

x = 0.9x+ 

15 

0.1x 

= 15 

x = 150 

87. a. Since the fund returns 8% compound 

annually, this is equivalent to a return of 2% 

each quarter. Defining a recursive 

sequence, we have: 

a0 = 0, an 

= 1.02an 

− 1+ 

500 

b. Insert the formulas in your graphing utility 

and use the table feature to find when the 

value of the account will exceed $100,000: 

86. a. p1 

= 0.9(250) + 15 = 240; 

p2 

= 0.9(240) + 15 = 231 

There are 240 tons of pollutants at the end of 

the first year, and 231 tons of pollutants at 

the end of the second year. 

b. Scrolling down the table, we display the 

pollutant levels for the next 20 years. 

In the 82nd quarter (approximately May 

2019) the value of the account will exceed 

$100,000 with a value of $101,810. 





c. Find the value of the account in 25 years or 

100 quarters: 

c. Enter the recursive formula in Y= and create 

the table: 

The value of the account will be $156,116.15. 

88. a. Since the fund returns 6% compounded 

annually, this is equivalent to a return of 

0.5% each month. Defining a recursive 

sequence, we have: 

a0 = 0, an 

= 1.005an 

− 1+ 

45 

b. Insert the formulas in your graphing utility 

and use the table feature to find when the 

value of the account will exceed $4,000: 

d. Scroll through the table: 

After 58 payments have been made, the 

balance is below $140,000. The balance is 

about $139,981. 

e. Scroll through the table: 

In the 74th month (February 2006) the value 

of the account will exceed $4,000 with a 

value of about $4,017.60. 

c. Find the value of the account in 16 years or 

192 months: 

The value of the account will be 

approximately $14,449.11. 

89. a. Since the interest rate is 6% per annum 

compounded monthly, this is equivalent to a 

rate of 0.5% each month. Defining a 

recursive sequence, we have: 

a0 = 150,000, an 

= 1.005an 

− 1− 

899.33 

b. 1.005(150,000) − 899.33 = $149,850.67 

The loan will be paid off at the end of 360 

months or 30 years. 

Total amount paid = (359)($899.33) + 

$890.65(1.005) = $323,754.57. 

f. The total interest expense is the difference 

of the total of the payments and the original 

loan: 323,754.57 − 150,000 = $173,754.57 

g. (a) Since the interest rate is 6% per annum 

compounded monthly, this is equivalent to 

a rate of 0.5% each month. Defining a 


a0 = 150,000, an 

= 1.005an 

− 1− 

999.33 

(b) 1.005(150,000) − 999.33 = $149,750.67 





(c) Enter the recursive formula in Y= and 

create the table: 

b. 

⎛ 0.065 ⎞ 

⎜1 + ⎟(120,000) −758.48 

= $119,891.52 




(d) Scroll through the table: 



balance is below $140,000. The 

balance is $139,894. 

(e) Scroll through the table: 

After 129 payments have been made, the 

balance is below $100,000. The balance is 

about $99,824. 


The loan will be paid off at the end of 

279 months or 23 years and 3 months. 

Total amount paid = (278)($999.33) + 

353.69(1.005) = $278,169.20 

(f) The total interest expense is the 

difference of the total of the payments 

and the original loan: 

278,169.20 − 150,000 = $128,169.20 

h. Yes, if they can afford the additional 

monthly payment. They would save 

$44,586.07 in interest payments by paying 

the loan off sooner. 

90. a. Since the interest rate is 6.5% per annum 

compounded at a rate of (6.5/12)% each 

month. Defining a recursive sequence, we 

have: 

a0 

= 120,000, 

⎛ 0.065 ⎞ 

an 

= ⎜1+ ⎟an 

− 1 −758.48 


The loan will be paid in the 360th month i.e. 

after 30 years. 

Total amount paid = (359)(758.48) + 

756.19(1+0.065/12) = $273,054.60. 



loan: 273,054.60 − 120,000 = $153,054.60 

g. (a) Since the interest rate is 6.5% per 

annum compounded monthly, this is 

equivalent to a rate of (6.5/12)% each 

month. Defining a recursive sequence, 

we have: 

a0 

= 120,000, 

⎛ 0.065 ⎞ 

an 

= ⎜1+ ⎟an 

− 1 −858.48 






(b) 

⎛ 0.065 ⎞ 

⎜1 + ⎟(120, 000) −858.48 


= $119,791.52 






balance is $99,831. 


The loan will be paid off at the end of 

262 months or 21 years and 10 months. 

Total amount paid = (261)(858.48) + 

(851.23)(1+0.065/12) = $224,919.12. 




224,919.12 − 120,000 = $104,919.12 

h. Yes, if they can afford the additional 

monthly payment. They would save 

$48,238.49 in interest payments by paying 

the loan off sooner. 

91. a1 a2 a3 a4 a5 

92. a. 

= 1, = 1, = 2, = 3, = 5, 

a6 = 8, a7 = 13, a8 = 21, an = an−1+ 

an−2 

a8 = a7 + a6 = 13+ 8 = 21 

After 7 months there are 21 mature pairs of 

rabbits. 

b. 

( 1+ 5) −( 1− 

5) 

1 1 

u1 = 

1 

2 5 

1+ 5− 1+ 

5 2 5 

= = = 1 

2 5 2 5 

u 

u 

= 

( 1+ 5) −( 1− 

5) 

2 2 

2 2 

= 

2 5 

1+ 2 5+ 5− 1+ 2 5−5 4 5 

= = = 1 

4 5 4 5 

n+ 

1 

+ u 

n 

n n n n 

+ 1 + 1 

( 1+ 5) −( 1− 5) ( 1+ 5) −( 1− 

5) 

+ 

n+ 

1 

n 

2 5 2 5 

+ 1 + 1 

( 1+ 5) −( 1− 5) + 2( 1+ 5) −2( 1− 

5) 

n n n n 

= 

n+ 

1 

2 5 

n 

n 

( 1+ 5) ⎡1+ 5 + 2⎤−( 1− 5) 

⎡1− 5+ 

2⎤ 

= 

⎣ ⎦ ⎣ ⎦ 

n+ 

1 

2 5 

n 

n 

( 1+ 5) ⎡3+ 5⎤−( 1− 5) 

⎡3− 

5⎤ 

= 

⎣ ⎦ ⎣ ⎦ 

n+ 

1 

2 5 

n+ 2 ( 3+ 5) 

n+ 

2 ( 3− 

5) 

( 1+ 5) 

− 

2 ( 1− 

5) 

2 

( 1+ 5) 

( 1− 

5) 

= 

n+ 

1 

2 5 

n+ 2 ( 3+ 5) 

n+ 

2 ( 3− 

5) 

( 1+ 5) 

−( 1− 

5) 

( 6+ 2 5) 

( 6−2 5) 

= 

n+ 

1 

2 5 

n+ 2 1 n+ 

2 1 

( 1+ 5) −( 1− 

5) 

= 

2 2 

n+ 

1 

2 5 

= 

n 

( 1+ 5) −( 1− 

5) 

= u n + 2 

+ 2 n+ 

2 

2 

n+ 

2 

5 

c. Since u = 1, u = 1, u = u + u , { u } 

1 2 n+ 2 n+ 

1 n n 

is the Fibonacci sequence. 





93. 1, 1, 2, 3, 5, 8, 13 

This is the Fibonacci sequence. 

94. a. u1 = 1, u2 = 1, u3 = 2, u4 = 3, u5 

= 5, 

u6 = 8, u7 = 13, u8 = 21, u9 = 34, u10 

= 55, 

u = 89 

b. 

11 

u2 1 u3 

2 u4 

3 

= = 1, = = 2, = = 1.5, 

u1 1 u2 1 u3 

2 

u5 5 u6 

8 

= ≈ 1.67, = = 1.6, 

u4 3 u5 

5 

u7 13 u8 

21 

= = 1.625, = ≈1.615, 

u6 8 u7 

13 

u9 34 u10 

55 

= ≈ 1.619, = ≈1.618, 

u8 21 u9 

34 

u11 

89 

= ≈1.618 

u 55 

10 

⎛ 

1+ 

5⎞ 

c. 1.618 ⎜The exact value is ⎟ 

⎝ 

2 ⎠ 

u1 1 u2 

1 u3 

2 

d. = = 1, = = 0.5, = ≈ 0.667, 

u2 1 u3 2 u4 

3 

u4 

3 u5 

5 

= = 0.6, = = 0.625, 

u5 5 u6 

8 

u6 8 u7 

13 

= ≈ 0.615, = ≈0.619, 

u7 13 u8 

21 

u8 21 u9 

34 

= ≈ 0.618, = ≈0.618, 

u9 34 u10 

55 

u10 

55 

= ≈0.618 

u 89 

11 

1.3 

c. f ( ) e 

1.3 = ≈ 3.669296668 

d. It will take n = 12 to approximate 

f 

1.3 

( 1.3) e 

96. a. ( 2.4) 

= correct to 8 decimal places. 

3 

( 2.4) 

∑ 

2.4 

f e − − 

− = ≈ 

b. ( 2.4) 

k = 0 

k! 

( −2.4) ( −2.4) ( −2.4) ( −2.4) 

+ + + 

0 1 2 3 

= 

0! 1! 2! 3! 

=−0.824 

2.4 ( 2.4) 

f e − − 

− = ≈ 

2.4 

c. f ( ) e − 

6 

∑ 

k = 0 

k! 

( −2.4) ( −2.4) ( −2.4) 

= + + ... + 

0 1 6 

0! 1! 6! 

= 0.1602688 

− 2.4 = ≈ 0.0907179533 

k 

k 

e. 0.618 

⎛ 

2 ⎞ 

⎜The exact value is ⎟ 

⎝ 

1 + 5 ⎠ 

d. It will take n = 17 to approximate 

f 

2.4 

( 2.4) e − 

− = correct to 8 decimal places. 

95. a. f ( 1.3) 

b. f ( 1.3) 

4 

k 

1.3 1.3 

∑ k! 

k = 0 

= e ≈ 

0 1 4 

1.3 1.3 1.3 

= + + ... + 

0! 1! 4! 

≈ 3.630170833 

7 

k 

1.3 1.3 

∑ k! 

k = 0 

= e ≈ 

0 1 7 

1.3 1.3 1.3 

= + + ... + 

0! 1! 7! 

≈ 3.669060828 





97. a. a 1 = 0.4 , 

2−2 

a 2 = 0.4 + 0.3⋅ 2 = 0.4 + 0.3 = 0.7 

3−2 

( ) 

( ) 

( ) 

( ) 

( ) 

( ) 

a 3 

= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 2 = 1.0 

4−2 

a 4 

= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 4 = 1.6 

5−2 

a 5 = 0.4 + 0.3⋅ 2 = 0.4 + 0.3 8 = 2.8 

6−2 

a 6 

= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 16 = 5.2 

7−2 

a 7 

= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 32 = 10.0 

8−2 

a 8 = 0.4 + 0.3⋅ 2 = 0.4 + 0.3 64 = 19.6 

The first eight terms of the sequence are 0.4, 

0.7, 1.0, 1.6, 2.8, 5.2, 10.0, and 19.6. 

b. Except for term 5, which has no match, 

Bode’s formula provides excellent 

approximations for the mean distances of 

the planets from the Sun. 

c. The mean distance of Ceres from the Sun is 

approximated by a 5 = 2.8 , and that of 

Uranus is a 8 = 19.6 . 

9−2 

d. = + ⋅ = + ( ) = 

10−2 

( ) 

a 9 

0.4 0.3 2 0.4 0.3 128 38.8 

a 10 = 0.4 + 0.3⋅ 2 = 0.4 + 0.3 256 = 77.2 

e. Pluto’s distance is approximated by a 9 , but 

no term approximates Neptune’s mean 

distance from the Sun. 

11−2 

f. ( ) 

a 11 

= 0.4 + 0.3⋅ 2 = 0.4 + 0.3 512 = 154 

According to Bode’s Law, the mean orbital 

distance of 2003 UB 313 will be 154 AU 

from the Sun. 

98. 5 

We begin with an initial guess of a 

0 

= 2 . 

1⎛ 

5⎞ 

a1 

= ⎜2+ ⎟= 

2.25 

2⎝ 

2⎠ 

1⎛ 

5 ⎞ 

a2 

= ⎜2.25 + ⎟≈2.236111111 

2⎝ 

2.25⎠ 

1⎛ 

5 ⎞ 

a3 

= ⎜2.236111111+ 

⎟ 

2 ⎝ 

2.236111111⎠ 

≈ 2.236067978 

1⎛ 

5 ⎞ 

a4 

= ⎜2.236067978 

+ 

⎟ 

2 ⎝ 

2.236067978 ⎠ 

≈ 2.236067977 

1⎛ 

5 ⎞ 

a5 

= ⎜2.236067977 

+ 

⎟ 

2 ⎝ 

2.236067977 ⎠ 

≈ 2.236067977 

For both a 

5 

and the calculator approximation, 

we obtain 5 ≈ 2.236067977 . 

99. 8 


0 

= 3 . 

1⎛ 8 ⎞ 1⎛ 8⎞ 

a1 = ⎜a0 

+ ⎟= ⎜3 + ⎟≈ 

2.833333333 

2⎝ 

a0 

⎠ 2⎝ 

3⎠ 

1⎛ 

8 ⎞ 

a2 = ⎜a1 

+ ⎟ 

2 ⎝ a1 

⎠ 

1⎛ 

8 ⎞ 

= ⎜2.833333333 

+ 

⎟ 

2 ⎝ 

2.2.833333333 ⎠ 

≈ 2.828431373 

1⎛ 

8 ⎞ 

a3 = ⎜a2 

+ ⎟ 

2 ⎝ a2 

⎠ 

1⎛ 

8 ⎞ 

= ⎜2.828431373 

+ 

⎟ 

2 ⎝ 

2.828431373 ⎠ 

≈ 2.828427125 

1⎛ 

8 ⎞ 

a4 = ⎜a3 

+ ⎟ 

2 ⎝ a3 

⎠ 

1 ⎛ 

8 ⎞ 

= ⎜ 2.828427125 + 

⎟ 

2 ⎝ 

2.828427125 ⎠ 

≈ 2.828427125 

1⎛ 

8 ⎞ 

a5 = ⎜a4 

+ ⎟ 

2 ⎝ a4 

⎠ 

1⎛ 

8 ⎞ 

= ⎜2.828427125 

+ 

⎟ 

2 ⎝ 

2.828427125 ⎠ 

≈ 2.828427125 

For both a 

5 


we obtain 8 ≈ 2.828427125 . 





100. 21 


0 

= 5 . 

1⎛ 

21⎞ 

a1 

= ⎜5+ ⎟= 

4.625 

2⎝ 

5 ⎠ 

1⎛ 

21 ⎞ 

a2 

= ⎜4.625 + ⎟≈ 

4.58277027 

2⎝ 

4.625⎠ 

1⎛ 

21 ⎞ 

a3 

= ⎜4.58277027 

+ 

⎟ 

2 ⎝ 

4.58277027 ⎠ 

≈ 4.582575699 

1⎛ 

21 ⎞ 

a4 

= ⎜4.582575699 

+ 

⎟ 

2 ⎝ 

4.582575699 ⎠ 

≈ 4.582575695 

1⎛ 

21 ⎞ 

a5 

= ⎜4.582575695 

+ 

⎟ 

2 ⎝ 

4.582575695 ⎠ 

≈ 4.582575695 

101. 89 


0 

= 9 . 

1⎛ 

89⎞ 

a1 

= ⎜5 + ⎟≈9.444444444 

2⎝ 

5 ⎠ 

1⎛ 

89 ⎞ 

a2 

= ⎜9.444444444 

+ 

⎟ 

2 ⎝ 

9.444444444 ⎠ 

≈ 9.433986928 

1⎛ 

89 ⎞ 

a3 

= ⎜9.433986928 

+ 

⎟ 

2 ⎝ 

9.433986928 ⎠ 

≈ 9.433981132 

1⎛ 

89 ⎞ 

a4 

= ⎜9.433981132 

+ 

⎟ 

2 ⎝ 

9.433981132 ⎠ 

≈ 9.433981132 

1⎛ 

89 ⎞ 

a5 

= ⎜9.433981132 

+ 

⎟ 

2 ⎝ 

9.433981132 

⎠ 

≈ 9.433981132 

For both a 

5 


we obtain 21 ≈ 4.582575695 . 

For both a 

5 


we obtain 89 ≈ 9.433981132 . 

_________________________________________________________________________________________________ 

102. To show that 1 2 3 ... ( n 1) 

( + 1) 

n n 

+ + + + − + n = 

2 

S = 1 + 2 + 3 + ... + n− 1 + n, we can reverse the order to get 

Let ( ) 

+ S = n+ ( n− ) + ( n− 

) 

1 2 +...+ 2 + 1, now add these two lines to get 

[ ] ⎡ ( ) ⎤ ⎡ ( ) ⎤ ⎡( ) ⎤ [ ] 

2S = 1+ n + ⎣2 + n− 1 ⎦+ ⎣3+ n− 2 ⎦+ ...... + ⎣ n− 1 + 2⎦+ n+ 

1 

So we have 2S = [ 1+ n] + [ 1+ n] + [ 1 + n] + .... + [ n+ 1] + [ n+ 1] = n⋅ [ n+ 

1] 

Thus, 2S 

= n( n+ 

1) 

n⋅ ( n+ 

1) 

S = 

103. Answers will vary. 

2 




Section 12.2: Arithmetic Sequences 


1. arithmetic 

2. False; the sum of the first and last terms equals 

twice the sum of all the terms divided by the 

number of terms. 

3. d = sn 

−sn−1 

= ( n+ 4) −( n− 1+ 4) = ( n+ 4) − ( n+ 

3) 

= n+ 4−n− 3= 

1 

The difference between consecutive terms is 

constant, therefore the sequence is arithmetic. 

s1 = 1+ 4= 5, s2 = 2+ 4= 6, s3 

= 3+ 4= 

7, 

s = 4+ 4= 

8 

4. 

5. 

6. 

4 

d = s −s 

n 

n−1 

( ) ( ) 

= ( n−5) −( n−1− 5) = n−5 − n−6 

= n−5− n+ 6= 

1 



s1 = 1− 5 =− 4, s2 = 2 − 5 =− 3, s3 

= 3 − 5 =−2, 

s = 4− 5= −1 

4 

d = an 

−an−1 

= ( 2n−5 ) −(2( n−1) −5) 

= ( 2n−5) −( 2n−2−5) 

= 2n−5− 2n+ 7 = 2 



a1 = 21 ⋅ − 5= − 3, a2 

= 22 ⋅ − 5= −1, 

a = 23 ⋅ − 5= 1, a = 24 ⋅ − 5= 

3 

3 4 

d = bn 

−bn−1 

= ( 3n+ 1 ) −(3( n− 1) + 1) 

= ( 3n+ 1) −( 3n− 3+ 

1) 

= 3n+ 1− 3n+ 2= 

3 



b1 = 31 ⋅ + 1= 4, b2 

= 32 ⋅ + 1= 

7, 

b = 3⋅ 3+ 1= 10, b = 3⋅ 4+ 1= 

13 

3 4 

7. 

8. 

d = c − c 

n 

n−1 

( n) 

n 

( 6 2n) ( 6 2n 

2) 

= 6−2 −(6−2( −1)) 

= − − − + 

= 6−2n− 6+ 2n− 2=−2 



c = 6−2⋅ 1= 4, c = 6−2⋅ 2= 

2, 

c 

1 2 

= 6−2⋅ 3= 0, c = 6−2⋅ 4=−2 

3 4 

d = d − d 

n 

n−1 

( n) 

n 

( 4 2n) ( 4 2n 

2) 

= 4−2 −(4−2( −1)) 

= − − − + 

= 4−2n− 4+ 2n− 2=−2 



d = 4−2⋅ 1= 2, d = 4−2⋅ 2= 

0, 

d 

1 2 

= 4−2⋅ 3= − 2, d = 4−2⋅ 4= −4 

3 4 

9. d = tn 

− tn−1 

⎛1 1 ⎞ ⎛1 1 ⎞ 

= ⎜ − n⎟−⎜ − ( n− 

1) ⎟ 

⎝2 3 ⎠ ⎝2 3 ⎠ 

⎛1 1 ⎞ ⎛1 1 1⎞ 

= ⎜ − n⎟−⎜ − n+ 

⎟ 

⎝2 3 ⎠ ⎝2 3 3⎠ 

1 1 1 1 1 1 

= − n− + n− = − 

2 3 2 3 3 3 



1 1 1 1 1 1 

t1 = − ⋅ 1 = , t2 

= − ⋅ 2 =− , 

2 3 6 2 3 6 

1 1 1 1 1 5 

t3 = − ⋅ 3 =− , t4 

= − ⋅ 4=− 

2 3 2 2 3 6 

10. d = tn 

− tn−1 

⎛2 1 ⎞ ⎛2 1 ⎞ 

= ⎜ + n⎟− ⎜ + ( n− 

1) ⎟ 

⎝3 4 ⎠ ⎝3 4 ⎠ 

⎛2 1 ⎞ ⎛2 1 1⎞ 

= ⎜ + n⎟− ⎜ + n− 

⎟ 

⎝3 4 ⎠ ⎝3 4 4⎠ 

2 1 2 1 1 1 

= + n− − n+ = 

3 4 3 4 4 4 



2 1 11 2 1 7 

t1 = + ⋅ 1 = , t2 

= + ⋅ 2 = , 

3 4 12 3 4 6 

2 1 17 2 1 5 

t3 = + ⋅ 3 = , t4 

= + ⋅ 4= 

3 4 12 3 4 3 





11. 

d = s −s 

n 

n−1 

n n−1 

( ) ( ) 

= ln 3 −ln 3 

nln ( 3) ( n 1) ln ( 3) 

( ln 3 )( n ( n 1 )) ( ln 3)( n n 1) 

= − − 

= − − = − + 

= ln 3 



s 

1 

= ln 3 = ln 3 , s 

2 

= ln 3 = 2ln 3 , 

s 

( ) ( ) ( ) ( ) 

3 4 

( ) ( ) s ( ) ( ) 

1 2 

= ln 3 = 3ln 3 , = ln 3 = 4ln 3 

3 4 

ln n ln( n−1) 

12. d s s e e n ( n ) 

= n − n−1 = − = − − 1 = 1 



ln1 ln 2 ln 3 

1 2 3 

ln 4 

4 = e = 4 

s = e = 1, s = e = 2, s = e = 3, 

s 

13. an 

= a1 + ( n−1) 

d 

= 2 + ( n −1)3 

= 2+ 3n 

−3 

= 3n 

−1 

a 51 = 351 ⋅ − 1= 

152 

14. an 

= a1 + ( n−1) 

d 

=− 2 + ( n −1)4 

=− 2+ 4n 

−4 

= 4n 

−6 

a 51 = 451 ⋅ − 6= 

198 

15. an 

= a1 + ( n−1) 

d 

= 5 + ( n −1)( −3) 

= 5− 3n 

+ 3 

= 8−3n 

a 51 = 8−3⋅ 51=− 

145 

16. an 

= a1 + ( n−1) 

d 

= 6 + ( n −1)( −2) 

= 6− 2n 

+ 2 

= 8−2n 

a 51 = 8 −2⋅ 51 =− 94 

17. 

an 

= a1 + ( n−1) 

d 

1 

= 0 + ( n −1) 2 

1 1 

= n − 

2 2 

1 

= ( n −1) 

2 

1 

a 51 = ( 51 − 1 ) = 25 

2 

18. an 

= a1 + ( n−1) 

d 

⎛ 1 ⎞ 

= 1 + ( n −1) 

⎜ − ⎟ 

⎝ 3 ⎠ 

1 1 

= 1− n + 3 3 

4 1 

= − n 

3 3 

4 1 4 51 47 

a 51 = − ⋅ 51 = − =− 

3 3 3 3 3 

19. an 

= a1 + ( n−1) 

d 

= 2 + ( n −1) 2 

= 2+ 2n− 2 = 2n 

a 51 = 51 2 

20. a a n d n ( n ) 

n = 1 + ( − 1) = 0 + ( −1) π= −1 

π 

a 51 = 51π−π= 50π 

21. a1 = 2, d = 2, an 

= a1+ ( n− 

1) d 

a 100 = 2 + (100 − 1)2 = 2 + 99(2) = 2 + 198 = 200 

22. a1 = − 1, d = 2, an 

= a1+ ( n− 

1) d 

a 80 = − 1 + (80 − 1)2 = − 1+ 79(2) =− 1+ 158 = 157 

23. a1 = 1, d =−2 − 1 =− 3, an 

= a1+ ( n− 

1) d 

a 90 = 1 + (90−1)( − 3) = 1+ 89( −3) 

= 1− 267= −266 

24. a1 = 5, d = 0 − 5 =− 5, an 

= a1+ ( n− 

1) d 

a 80 = 5 + (80−1)( − 5) = 5+ 79( −5) 

= 5 − 395 = −390 





5 1 

25. a1 = 2, d = − 2 = , an 

= a1+ ( n− 

1) d 

2 2 

1 83 

a 80 = 2 + (80− 1) = 2 2 

26. a1 

= 2 5, d = 4 5− 2 5 = 2 5, 

an 

= a1 

+ ( n−1) 

d 

a 70 = 2 5 + (70−1)2 5 

= 2 5+ 

69 2 5 

( ) 

= 2 5+ 

138 5 

= 140 5 

27. 8 1 20 1 

a = a + 7d = 8 a = a + 19d 

= 44 

Solve the system of equations by subtracting the 

first equation from the second: 

12d 

= 36 ⇒ d = 3 

a1 

= 8− 7(3) = 8− 21=−13 

Recursive formula: a1 =− 13 an 

= an 

− 1+ 

3 

nth term: an 

= a1 + ( n−1) 

d 

=− 13 + ( n −1)( 3) 

=− 13 + 3n 

−3 

= 3n 

−16 

28. 4 1 20 1 

a = a + 3d = 3 a = a + 19d 

= 35 



16d 

= 32 ⇒ d = 2 

a1 

= 3− 3(2) = 3− 6=−3 


= an 

− 1+ 

2 

nth term: an 

= a1 + ( n−1) 

d 

=− 3+ ( n −1)( 2) 

=− 3+ 2n 

−2 

= 2n 

−5 

29. 9 1 15 1 

a = a + 8d = − 5 a = a + 14d 

= 31 



6d 

= 36⇒ d = 6 

a1 

=−5 − 8(6) =−5 − 48 =−53 


= an 

− 1+ 

6 

nth term: an 

= a1 + ( n−1) 

d 

=− 53 + ( n −1)( 6) 

=− 53 + 6n 

−6 

= 6n 

−59 

30. 8 1 18 1 

a = a + 7d = 4 a = a + 17d 

=− 96 



10d 

= −100 ⇒ d =−10 

a = 4 −7( − 10) = 4 + 70 = 74 

1 

Recursive formula: a1 = 74 an 

= an 

− 1− 

10 

a = a + n− 

d 

nth term: n 1 ( 1) 

= 74 + ( n −1)( −10) 

= 74 − 10n 

+ 10 

= 84 −10n 

31. 15 1 40 1 

a = a + 14d = 0 a = a + 39d 

= − 50 



25d 

= −50 ⇒ d = −2 

a =−14( − 2) = 28 

1 


= an 

− 1− 

2 

a = a + n− 

d 


= 28 + ( n −1)( −2) 

= 28 − 2n 

+ 2 

= 30 −2n 

32. 5 1 13 1 

a = a + 4d =− 2 a = a + 12d 

= 30 



8d 

= 32⇒ d = 4 

a = −2− 4(4) = −18 

1 


= an 

− 1+ 

4 

a = a + n− 

d 


=− 18 + ( n −1)( 4) 

=− 18 + 4n 

−4 

= 4n 

−22 

33. 14 1 18 1 

a = a + 13d = − 1 a = a + 17d 

=− 9 



4d 

= −8⇒ d = −2 

a = −1−13( − 2) = − 1+ 26= 

25 

1 


= an 

− 1− 

2 

a = a + n− 

d 


= 25 + ( n −1)( −2) 

= 25 − 2n 

+ 2 

= 27 −2n 





34. 12 1 18 1 

a = a + 11d = 4 a = a + 17d 

= 28 



6d 

= 24⇒ d = 4 

a = 4− 11(4) = 4− 44= −40 

1 


= an 

− 1+ 

4 

a = a + n− 

d 


=− 40 + ( n −1)( 4) 

=− 40 + 4n 

−4 

= 4n 

−44 

n n n 

1 n 1 2 1 2 

2 2 2 

2 

35. S = ( a + a ) = ( + ( n− )) = ( n) = n 

n 

n n 

1 n 2 2 1 

2 2 

2 

36. S = ( a + a ) = ( + n) = n+ n = n( n+ 

) 

n 

n n n 

1 n 7 2 5 9 5 

2 2 2 

37. S = ( a + a ) = ( + ( + n) 

) = ( + n) 

n 

n n 

2 

1 n 

2 

n 

2 

= ( 4n− 6) 

= 2n −3n 

2 

= n − 

38. S = ( a + a ) = ( − 1+ ( 4n−5) 

) 

n 

( 2n 

3) 

39. a1 = 2, d = 4 − 2 = 2, an 

= a1+ ( n− 

1) d 

70 = 2 + ( n −1)2 

70 = 2 + 2n 

−2 

70 = 2n 

n = 35 

n 35 35 

Sn 

= ( a1 

+ an) = ( 2+ 70) = ( 72) 

= 1260 

2 2 2 

40. a1 = 1, d = 3 − 1 = 2, an 

= a1+ ( n− 

1) d 

59 = 1 + ( n −1)2 

59 = 1+ 2n 

−2 

60 = 2n 

n = 30 

n 30 

Sn 

= ( a1 

+ an) = ( 1 + 59 ) = 15 ( 60 ) = 900 

2 2 

41. a1 = 5, d = 9 − 5 = 4, an 

= a1+ ( n− 

1) d 

49 = 5 + ( n −1) 

4 

49 = 5 + 4n 

−4 

48 = 4n 

n = 12 

n 12 

Sn 

= ( a1 

+ an) = ( 5 + 49 ) = 6 ( 54 ) = 324 

2 2 

42. a1 = 2, d = 5 − 2 = 3, an 

= a1+ ( n− 

1) d 

41 = 2 + ( n −1) 

3 

41 = 2 + 3n 

−3 

42 = 3n 

n = 14 

n 14 

Sn 

= ( a1 

+ an) = ( 2 + 41 ) = 7 ( 43 ) = 301 

2 2 

43. a 1 = 73 , 78 73 5 

d = − = , 

n 

= 

1 

+ ( − 1) 

( n )( ) 

( n ) 

558 = 73 + −1 5 

485 = 5 −1 

97 = n −1 

98 = n 

n 98 

Sn 

= a1 

+ an 

= 73 + 558 

2 2 

= 49 631 = 30,919 

( ) ( ) 

( ) 

44. a 1 = 7 , 1 7 6 

a a n d 

d = − = − , 

n = 1 + ( − 1) 

( n )( ) 

( n ) 

− 299 = 7 + −1 −6 

− 306 = −6 −1 

51 = n −1 

52 = n 

n 52 

Sn 

= a1 

+ an 

= 7 + − 299 

2 2 

= 26 − 292 = −7592 

a a n d 

( ) 

( ) ( ) 

( ) 

45. a 1 = 4 , 4.5 4 0.5 

d = − = , 

n 

= 

1 

+ ( − 1) 

( n )( ) 

( n ) 

100 = 4 + −1 0.5 

a a n d 

96 = 0.5 −1 

192 = n −1 

193 = n 

n 193 193 

Sn 

= a1 

+ an 

= 4+ 100 = 104 

2 2 2 

= 10,036 

( ) ( ) ( ) 





1 1 

46. a 1 = 8 , d = 8 − 8= , an 

= a1 + ( n− 

1) 

d 

4 4 

⎛1 

⎞ 

50 = 8 + ( n −1) 

⎜ ⎟ 

⎝4 

⎠ 

1 

42 = ( n −1) 

4 

168 = n −1 

169 = n 

n 169 169 

Sn 

= ( a1 

+ an) = ( 8 + 50) = ( 58) 

= 4901 

2 2 2 

47. a 

1 

= 21 () − 5= − 3, ( ) 

a 

80 

= 280− 5= 

155 

80 

S 

80 

= ( − 3+ 155) = 40( 152) 

= 6080 

2 

48. a 1 = 3− 2() 

1 = 1, ( ) 

a 90 = 3 − 2 90 =− 177 

( ( )) ( ) 

90 

S 

90 

= 1 + − 177 = 45 − 176 = − 7920 

2 

1 11 1 

49. a 

1 

= 6− () 1 = , a 

100 

= 6 − ( 100) 

= − 44 

2 2 

2 

100 ⎛11 ⎞ ⎛ 77 ⎞ 

S100 

= ⎜ + ( − 44) 

⎟= 50⎜− ⎟= −1925 

2 ⎝ 2 ⎠ ⎝ 2 ⎠ 

1 1 5 

50. 1 () 1 

3 2 6 

1 1 163 

a = + = 

3 2 6 

a = + = , 80 ( 80) 

S 

80 

80 ⎛5 163 ⎞ 

= ⎜ + ⎟= 40( 28) 

= 1120 

2 ⎝6 6 ⎠ 

d = − = , an 

= a1 + ( n− 

1) 

d 

a 


= 14 + ( 120 − 1)( 2) = 14 + 119( 2) 

= 252 

51. a 1 = 14 , 16 14 2 


S 


= ( 14 + 252 ) = 60 ( 266 ) = 15,960 

2 

d =− − =− , an 

= a1 + ( n− 

1) 

d 

a 

46 

= 2+ ( 46−1)( − 3) = 2+ ( 45)( − 3) 

= − 133 

52. a 1 = 2 , 1 2 3 

( ( )) ( ) 

46 

S 

46 

= 2 + − 133 = 23 − 131 = − 3013 

2 

53. Find the common difference of the terms and 

solve the system of equations: 

(2x + 1) − ( x+ 3) = d ⇒ x− 2 = d 

(5x + 2) − (2x+ 1) = d ⇒ 3x+ 1 = d 

3x+ 1= x−2 

2x 

=−3 

3 

x =− 

2 

54. Find the common difference of the terms and 

solve the system of equations: 

(3x + 2) − (2 x) = d ⇒ x+ 2 = d 

(5x + 3) − (3x+ 2) = d ⇒ 2x+ 1 = d 

2x+ 1= x+ 

2 

x = 1 

55. a 1 = 11 , d = 3 , S = 1092 

n 

( ) 

an 

= a1 + ( n− 1) d = 11+ n− 1 3= 3n+ 

8 

n 

Sn 

= ( a1 

+ an) 

2 

n 

1092 = ⎡11 (3 8) 

2 

⎣ + n + ⎤⎦ 

n 

1092 = (3n 

+ 19) 

2 

2184 = n(3n+ 

19) 

2 

0 = 3n 

+ 19n−2184 

0 = (3n+ 91)( n−24) 

91 

n =− or n = 24 

3 

Disregard the negative result. To obtain a sum 

of 1092, add 24 terms. 

56. a 1 = 78 , d = − 4 , S = 702 

n 

an 

= a1 + ( n− 1) d = 78 + ( n−1)( − 4) = − 4n+ 

82 

n 

Sn 

= ( a1 

+ an) 

2 

n 

702 = ⎡78 ( 4 82) 

2 

⎣ + − n + ⎤⎦ 

n 

702 = ( − 4n 

+ 160) 

2 

2 

0 =− 2n 

+ 80n−702 

2 

0= n − 40n+ 

351 

0 = ( n−27)( n−13) 

n= 27 or n= 

13 

To obtain a sum of 702, add 13 terms or 27 terms. 





57. The total number of seats is: 

S = 25 + 26 + 27 + + ( 25 + 29() 

1 ) 

This is the sum of an arithmetic sequence with 

d = 1, a1 

= 25, and n= 30 . 

Find the sum of the sequence: 

30 

S 30 = [ 2(25) + (30 − 1)(1) ] 

2 

= 15(50 + 29) = 15(79) 

= 1185 

There are 1185 seats in the theater. 

58. The total number of seats is: 

( ( ( ))) 

S = 15 + 17 + 19 + + 15 + 39 2 

This is the sum of an arithmetic sequence with 

d = 2, a1 

= 15, and n = 40 . 

Find the sum of the sequence: 

40 

S 40 = [ 2(15) + (40 − 1)(2) ] 

2 

= 20(30 + 78) = 20(108) 

= 2160 

The corner section has 2160 seats. 

59. The lighter colored tiles have 20 tiles in the 

bottom row and 1 tile in the top row. The 

number decreases by 1 as we move up the 

triangle. This is an arithmetic sequence with 

a1 = 20, d =− 1, and n = 20 . Find the sum: 

20 

S = [ 2(20) + (20 − 1)( − 1) ] 

2 

= 10(40 − 19) = 10(21) 

= 210 

There are 210 lighter tiles. 

The darker colored tiles have 19 tiles in the 

bottom row and 1 tile in the top row. The 

number decreases by 1 as we move up the 

triangle. This is an arithmetic sequence with 

a1 = 19, d =− 1, and n = 19 . Find the sum: 

19 

S = [ 2(19) + (19 − 1)( − 1) ] 

2 

19 19 

= (38 − 18) = (20) = 190 

2 2 

There are 190 darker tiles. 

60. The number of bricks required decreases by 2 on 

each successive step. This is an arithmetic 

sequence with a1 = 100, d =− 2, and n= 30 . 

a. The number of bricks for the top step is: 

a30 = a1 + ( n− 1) d = 100 + (30 −1)( −2) 

= 100 + 29( − 2) = 100 −58 

= 42 

42 bricks are required for the top step. 

b. The total number of bricks required is the 

sum of the sequence: 

30 

S = [ 100 + 42 ] = 15(142) = 2130 

2 

2130 bricks are required to build the 

staircase. 

61. The air cools at the rate of 5.5° F per 1000 feet. 

Since n represents thousands of feet, we have 

d = − 5.5 . The ground temperature is 67° F so 

we have T 1 = 67 − 5.5 = 61.5 . Therefore, 

T = 61.5 + n−1 −5.5 

{ } 

{ n} ( )( ) 

= { − 5.5n+ 67 } or { 67 −5.5n} 

After the parcel of air has risen 5000 feet, we 

have T 5 = 61.5 + ( 5 −1)( − 5.5) 

= 39.5 . 

The parcel of air will be 39.5° F after it has risen 

5000 feet. 

62. If we treat the length of each rung as the term of 

an arithmetic sequence, we have 1 49 a = , 

d = − 2.5 , and a = 24 . 

n 

n 

( ) 

( n )( ) 

( n ) 

a = a + n− 

d 

1 

1 

24 = 49 + −1 −2.5 

− 25 =−2.5 −1 

10 = n −1 

11 = n 

Therefore, the ladder contains 11 rungs. 

To find the total material required for the rungs, 

we need the sum of their lengths. Since there are 

11 rungs, we have 

11 11 

S 

11 

= ( 49 + 24) = ( 73) 

= 401.5 

2 2 

It would require 401.5 feet of material to 

construct the rungs for the ladder. 




Section 12.3: Geometric Sequences; Geometric Series 

63. a 1 = 35 , 37 35 2 

d = − = , 

n 

= 

1 

+ ( − 1) 

a 27 = 35 + ( 27 − 1)( 2) = 35 + 26( 2) 

= 87 

a a n d 

27 27 

S 

27 

= ( 35 + 87) = ( 122) 

= 1647 

2 2 

The amphitheater has 1647 seats. 

64. Find n in an arithmetic sequence with 

a1 = 10, d = 4, S n = 2040 . 

n 

Sn 

= [ 2 a1 

+ ( n−1) 

d] 

2 

n 

2040 = [ 2(10) + ( n −1)4] 

2 

4080 = n[ 20 + 4n−4] 

4080 = n(4n+ 

16) 

2 

4080 = 4n 

+ 16n 

2 

1020 = n + 4n 

2 

n + 4n− 1020= 

0 

( n+ 34)( n− 30) = 0 ⇒ n =− 34 or n= 

30 

There are 30 rows in the corner section of the 

stadium. 

65. The yearly salaries form an arithmetic sequence 

with a1 = 35,000, d = 1400, S n = 280,000 . 

Find the number of years for the aggregate salary 

to equal $280,000. 

n 

Sn 

= [ 2 a1 

+ ( n−1) 

d] 

2 

n 

280,000 = [ 2(35,000) + ( n −1)1400] 

2 

280,000 = n[ 35,000 + 700n−700] 

280,000 = n(700n+ 

34,300) 

2 

280,000 = 700 n + 34,300 n 

2 

400 = n + 49 n 

2 

n + 49 n− 400 = 0 

2 

− 49 ± 49 −4(1)( −400) 

n = 

2(1) 

− 49 ± 4001 − 49 ± 63.25 

= ≈ 

2 2 

n≈7.13 or n ≈− 56.13 

It takes about 8 years to have an aggregate salary 

of at least $280,000. The aggregate salary after 8 

years will be $319,200. 


67. Answers will vary. Both increase (or decrease) at 

a constant rate, but the domain of an arithmetic 

sequence is the set of natural numbers while the 

domain of a linear function is the set of all real 

numbers. 


1. geometric 

2. 

a 

1− r 

3. divergent series 

4. True 

5. False; the common ratio can be positive or 

negative (or 0, but this results in a sequence of 

only 0s). 

6. True 

7. 

8. 

n+ 

1 

3 n+− 

1 n 

r = = 3 = 3 

n 

3 

The ratio of consecutive terms is constant, 

therefore the sequence is geometric. 

1 2 

1 = = s2 

= = 

3 4 

3 s4 

s 

s 

3 3, 3 9, 

= 3 = 27, = 3 = 81 

n+ 

1 

( −5) 

n+− 

1 n 

r = = ( − 5) = −5 

n 

( −5) 



1 2 

1 = − =− s2 

= − = 

3 4 

3 s4 

s 

s 

( 5) 5, ( 5) 25, 

= ( − 5) = − 125, = ( − 5) = 625 

n+ 

1 

⎛1 

⎞ 

−3 

⎜ ⎟ n+− 

1 n 

2 ⎛1⎞ 

1 

9. r = 

⎝ ⎠ 

= 

n ⎜ ⎟ = 

⎛1 

⎞ ⎝2⎠ 

2 

−3 ⎜ ⎟ 

⎝ 2 ⎠ 



1 2 

⎛1⎞ 3 ⎛1⎞ 

3 

a1 =− 3 ⎜ ⎟ =− , a2 

=− 3 ⎜ ⎟ =− , 

⎝2⎠ 2 ⎝2⎠ 

4 

3 4 

⎛1⎞ 3 ⎛1⎞ 

3 

a3 = − 3 ⎜ ⎟ = − , a4 

=− 3⎜ ⎟ =− 

⎝2⎠ 8 ⎝2⎠ 

16 





10. 

11. 

12. 

13. 

n+ 

1 

⎛5 

⎞ 

⎜ ⎟ 

n+− 

1 n 

2 ⎛5⎞ 

5 

r = 

⎝ ⎠ 

= 

n ⎜ ⎟ = 

⎛5 

⎞ ⎝2⎠ 

2 

⎜ ⎟ 

⎝2 

⎠ 



1 2 

⎛5⎞ 5 ⎛5⎞ 

25 

b1 = ⎜ ⎟ = , b2 

= ⎜ ⎟ = , 

⎝2⎠ 2 ⎝2⎠ 

4 

3 4 

⎛5 ⎞ 125 ⎛5 ⎞ 625 

b3 = ⎜ ⎟ = , b4 

= ⎜ ⎟ = 

⎝2⎠ 8 ⎝2⎠ 

16 

⎛ 

n+− 

11 

2 ⎞ 

⎜ 

4 ⎟ n 

2 n−( n−1) 

r = 

⎝ ⎠ 

= = 2 = 2 

n−1 

n−1 

⎛2 

⎞ 2 

⎜ 

4 ⎟ 

⎝ ⎠ 



11 − 0 

2 2 −2 

1 

c1 = = = 2 = , 

2 

4 2 4 

2−1 1 

2 2 −1 

1 

c2 = = = 2 = , 

2 

4 2 2 

3−1 2 

2 2 

3 2 

c 

c 

= = = 1, 

4 2 

4−1 3 

2 2 

4 2 

= = = 2 

4 2 

⎛ 

n+ 

1 

3 ⎞ 

⎜ 

9 ⎟ n+ 

1 

3 n+− 

1 n 

r = 

⎝ ⎠ 

= = 3 = 3 

n n 

⎛3 

⎞ 3 

⎜ 

9 ⎟ 

⎝ ⎠ 



1 2 

1 d2 

3 1 3 9 

d = = , = = = 1, 

9 3 9 9 

3 4 

3 27 3 81 

d3 = = = 3, d4 

= = = 9 

9 9 9 9 

⎛n+ 

1⎞ 

⎜ ⎟ 

3 

⎛n+ 

1 n⎞ 

⎝ ⎠ ⎜ − ⎟ 

⎝ 3 3⎠ 

⎛n 

⎞ 

⎜ ⎟ 

⎝3 

⎠ 

2 

r = = 2 = 2 

1/3 

2 



1/3 2/3 3/3 4/3 

1 = 2 , 2 = 2 , 3 = 2 = 2, 4 = 2 

e e e e 

14. 

15. 

16. 

17. 

18. 

19. 

20. 

2( n+ 

1) 

3 

2n+ 2−2n 

2 

r = = 3 = 3 = 9 

2n 

3 



21 ⋅ 

22 ⋅ 4 

f1 = 3 = 9, f2 

= 3 = 3 = 81, 

23 ⋅ 6 24 ⋅ 8 

f = 3 = 3 = 729, f = 3 = 3 = 6561 

3 4 

⎛ 

n+− 

11 

3 ⎞ 

⎜ n+ 

1 

2 ⎟ n n 

3 2 

r = 

⎝ ⎠ 

= ⋅ 

n 1 n− 1 n+ 

1 

⎛ 

− 

3 ⎞ 3 2 

⎜ n 

2 ⎟ 

⎝ ⎠ 

n−( n−1) n− ( n+ 1) −1 

3 

= 3 ⋅ 2 = 3⋅ 2 = 

2 



11 − 0 21 − 1 

1 t 

1 2 2 2 

3 3 1 3 3 3 

t = = = , = = = , 

2 2 2 2 2 4 

3−1 2 4−1 3 

3 3 9 3 3 27 

t3 = = = , t 

3 3 4 = = = 

4 4 

2 2 8 2 2 16 

⎛ 

n+ 

1 

2 ⎞ 

⎜ n+− 11 1 1 

3 ⎟ n− n+ 

3 2 

r = 

⎝ ⎠ 

= ⋅ 

n n n 

⎛ 2 ⎞ 3 2 

⎜ n−1 

3 ⎟ 

⎝ ⎠ 

n−− 1 n n+− 1 n −1 

2 

= 3 ⋅ 2 = 3 ⋅ 2= 

3 



1 2 

1 11 − 0 

u2 

21 − 

2 2 2 2 4 

u = = = = 2, = = , 

3 3 1 3 3 

3 4 

2 8 8 2 16 16 

u3 = = = , u 

3−1 2 4 = = = 

4−1 3 

3 3 9 3 3 27 

a 

a 

5 

n 

a 

a 

5 

n 

a 

a 

5 

n 

n 

5−1 4 

= 23 ⋅ = 23 ⋅ = 281 ⋅ = 162 

1 

23 n − 

= ⋅ 

5−1 4 

=−2⋅ 4 =−2⋅ 4 =−2⋅ 256 =−512 

1 

24 n − 

=− ⋅ 

5−1 4 

= 5( − 1) = 5( − 1) = 5⋅ 1 = 5 

1 

5(1) n − 

= ⋅ − 

5−1 4 

a5 

= 6( − 2) = 6( − 2) = 6⋅ 16 = 96 

1 

a 6( 2) n − 

= ⋅ − 





21. 

22. 

a 

a 

a 

a 

5 

n 

5 

n 

5−1 4 

⎛1⎞ ⎛1⎞ 

= 0⋅ ⎜ ⎟ = 0⋅ ⎜ ⎟ = 0 

⎝2⎠ ⎝2⎠ 

n−1 

⎛1 

⎞ 

= 0⋅ ⎜ ⎟ = 0 

⎝2 

⎠ 

5−1 4 

⎛ 1⎞ ⎛ 1⎞ 

1 

= 1⋅⎜− ⎟ = 1⋅⎜− ⎟ = 

⎝ 3⎠ ⎝ 3⎠ 

81 

n−1 n−1 

⎛ 1⎞ ⎛ 1⎞ 

= 1⋅⎜− ⎟ = ⎜− 

⎟ 

⎝ 3⎠ ⎝ 3⎠ 

− 

23. a5 

( ) ( ) 

n−1 

n 

an 

= 2⋅ ( 2) = ( 2) 

24. 

a 

a 

5 

n 

5 1 4 

= 2⋅ 2 = 2⋅ 2 = 2⋅ 4= 

4 2 

5−1 4 

⎛1⎞ ⎛1⎞ 

= 0⋅ ⎜ ⎟ = 0⋅ ⎜ ⎟ = 0 

⎝π⎠ ⎝π⎠ 

n−1 

⎛1 

⎞ 

= 0⋅ ⎜ ⎟ = 0 

⎝π 

⎠ 

1 

25. a1 

= 1, r = , n = 7 

2 

a 

7 

7−1 6 

⎛1⎞ ⎛1⎞ 

1 

= 1⋅ ⎜ ⎟ = ⎜ ⎟ = 

⎝ 2 ⎠ ⎝ 2 ⎠ 64 

26. a1 

= 1, r = 3, n = 8 

27. 

a 

8 

1 

9 

8−1 7 

= 1⋅ 3 = 3 = 2187 

a = 1, r =− 1, n = 9 

a 

− 

( ) ( ) 

9 1 8 

= 1⋅ − 1 = − 1 = 1 

10 

n 

32. a 1 = 5 , r = = 2 , an 

= a1 

⋅ r − 

5 

1 

52 n − 

= ⋅ 

a n 

1 1 

n 

33. a 1 = − 3 , r = =− , an 

= a1 

⋅ r − 

−3 3 

n−1 n−2 

⎛ 1⎞ ⎛ 1⎞ 

a n =−3⎜− ⎟ = ⎜− 

⎟ 

⎝ 3⎠ ⎝ 3⎠ 

1 

n 

34. a 1 = 4 , r = , an 

= a1 

⋅ r − 

4 

n−1 n−2 

⎛1⎞ ⎛1⎞ 

a n = 4⎜ ⎟ = ⎜ ⎟ 

⎝ 4 ⎠ ⎝ 4 ⎠ 

35. 

36. 

n 1 

an 

= a1 

⋅ r − 

243 = a ⋅ −3 

1 

1 

( ) 

5 

( ) 

243 = a1 

−3 

243 =−243a 

− 1 = a 

Therefore, 

a = a ⋅ r − 

1 

a n 

n 1 

n 1 

2−1 

⎛1 

⎞ 

7 = a1 

⎜ ⎟ 

⎝3 

⎠ 

1 

7 = a1 

3 

21 = a 

1 

Therefore, 

a n 

6−1 

=−− ( 3) n 

−1 

n 

⎛1 

⎞ 

= 21⎜ ⎟ 

⎝ 3 ⎠ 

−1 

. 

1 

1 

1 

28. 

29. 

30. 

a =− 1, r =− 2, n= 

10 

a 

1 

10 

1 

8 

− 

( ) ( ) 

10 1 9 

=−1⋅ − 2 = −1⋅ − 2 =−1( − 512) = 512 

a = 0.4, r = 0.1, n= 

8 

a 

1 

7 

− 

( ) ( ) 

8 1 7 

= 0.4⋅ 0.1 = 0.4 0.1 = 0.00000004 

a = 0.1, r = 10, n= 

7 

a 

7−1 

( ) 

= 0.1⋅ 10 = 0.1 10 = 100,000 

14 

n 

31. a 1 = 7 , r = = 2 , an 

= a1 

⋅ r − 

7 

1 

7 2 n − 

= ⋅ 

a n 

6 

1 

37. 

2 1 

4−1 3 

a4 a1⋅ 

r r 

= = = r 

2−1 

a a ⋅ r r 

2 1575 

r = = 225 

7 

r = 225 = 15 

a = a ⋅r 

n 

1 

1 

1 

n−1 

7 = a ⋅15 

7 = 15a 

7 

a1 

= 

15 

Therefore, 

2−1 

a n 

2 

7 15 

n−1 7 15 

n−2 

= ⋅ = ⋅ . 

15 





38. 

39. 

3 1 

6−1 5 

a6 a1 

⋅ r r 

= = = r 

3−1 2 

a a ⋅ r r 

r 

1 

3 81 

1 

3 

1 1 

= = ⋅ 3 = 

81 27 

1 1 

r = 3 = 

27 3 

n−1 

a = a ⋅r 

n 1 

3−1 

1 ⎛1⎞ 

= a1 

⋅ ⎜ ⎟ 

3 ⎝3⎠ 

1 1 

= a1 

3 9 

3 = a 

1 

Therefore, 

a n 

3 

n−1 n−2 

⎛1⎞ ⎛1⎞ 

= 3⎜ ⎟ = ⎜ ⎟ 

⎝ 3 ⎠ ⎝ 3 ⎠ 

1 

a1 

= , r = 2 

4 

⎛ 

n 

n 

1−r 

⎞ 1⎛1−2 ⎞ 1 

Sn 

= a1 

⎜ 1 2 

1 r ⎟ 

= 4⎜ 1 2 ⎟ 

= − − 

⎝ − ⎠ ⎝ − ⎠ 4 

1 

( 2 

n 

= − 1 ) 

4 

. 

n 

( ) 

42. 

a = 4, r = 3 

S 

1 

n 

⎛ 

n n n 

1−r 

⎞ ⎛1−3 ⎞ ⎛1−3 

⎞ 

= a1 

⎜ = 4 = 4 

1−r 

⎟ ⎜ 1−3 ⎟ ⎜ −2 

⎟ 

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 

n n 

( ) ( ) 

=−21− 3 = 23 −1 

43. a1 

= − 1, r = 2 

S 

n 

⎛ 

n 

n 

1−r 

⎞ ⎛1−2 

⎞ 

= a1 

⎜ =− 1 = 1−2 

1−r 

⎟ ⎜ 1−2 

⎟ 

⎝ ⎠ ⎝ ⎠ 

3 

44. a1 

= 2, r = 

5 

S 

n 

⎡ 

n 

n 

⎛3⎞ ⎤ ⎡ ⎛3⎞ 

⎤ 

n ⎢1− 

⎥ ⎢1− 

⎥ 

⎛1− r ⎞ ⎜ ⎟ ⎜ ⎟ 

5 5 

= a1 

2 

⎝ ⎠ 

2 

⎝ ⎠ 

⎢ ⎥ ⎢ ⎥ 

⎜ 1 r ⎟ 

= = 

⎢ 3 ⎥ ⎢ 2 ⎥ 

⎝ − ⎠ 1 

⎛ ⎞ 

⎢ − 5 

⎥ ⎢ ⎜ ⎟ ⎥ 

⎢⎣ ⎥⎦ ⎢⎣ ⎝ 5 ⎠ ⎥⎦ 

⎡ 

n 

⎛3 

⎞ ⎤ 

= 51 ⎢ −⎜ 

⎟ ⎥ 

⎢⎣ 

⎝5 

⎠ ⎥⎦ 

45. Using the sum of the sequence feature: 

n 

40. 

3 1 

a1 

= = , r = 3 

9 3 

⎛ 

n n n 

1−r 

⎞ 1⎛1−3 ⎞ 1⎛1−3 

⎞ 

Sn 

= a1 

⎜ 1 r ⎟ 

= 3⎜ 1 3 ⎟ 

= 

3⎜ 2 ⎟ 

⎝ − ⎠ ⎝ − ⎠ ⎝ − ⎠ 

1 n 1 n 

=− ( 1− 3 ) = ( 3 −1) 

6 6 


2 2 

41. a1 

= , r = 

3 3 

S 

n 

⎡ 

n 

⎛2 

⎞ ⎤ 

n ⎢1− 

⎥ 

⎛1− r ⎞ 2 

⎜ ⎟ 

3 

= a 

⎝ ⎠ 

1 

⎢ ⎥ 

⎜ 1 r ⎟ 

= 

3⎢ 2 ⎥ 

⎝ − ⎠ 

⎢ 1− 

3 

⎥ 

⎢⎣ 

⎥⎦ 

⎡ 

n 

⎛2 

⎞ ⎤ 

⎢1− 

2 

⎜ ⎟ ⎥ 

n 

3 ⎡ ⎛2⎞ 

⎤ 

= ⎢ ⎝ ⎠ ⎥ = 21 ⎢ − ⎥ 

3⎢ 1 ⎥ ⎜ ⎟ 

⎢ ⎝3⎠ 

⎥ 

⎢ 

⎣ ⎦ 

3 

⎥ 

⎢⎣ 

⎥⎦ 









1 

51. a1 

= 1, r = 

3 

Since r < 1, the series converges. 

S 

∞ 

a1 

1 1 3 

= = = = 

1− r ⎛ 1⎞ ⎛2⎞ 

2 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 3 ⎠ ⎝ 3 ⎠ 

2 

52. a1 

= 2, r = 

3 


S 

∞ 

a1 

2 2 

= = = = 6 

1− r ⎛ 2⎞ ⎛1⎞ 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 3 ⎠ ⎝ 3 ⎠ 

1 

53. a1 

= 8, r = 

2 


S 

∞ 

a1 

8 8 

= = = = 16 

1− r ⎛ 1⎞ ⎛1⎞ 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 2 ⎠ ⎝ 2 ⎠ 

1 

54. a1 

= 6, r = 

3 


S 

∞ 

a1 

6 6 

= = = = 9 

1− r ⎛ 1⎞ ⎛2⎞ 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 3 ⎠ ⎝ 3 ⎠ 

1 

55. a1 

= 2, r =− 

4 


a1 

2 2 8 

S∞ 

= = = = 

1− r ⎛ ⎛ 1 ⎞⎞ ⎛5 

⎞ 5 

⎜1−⎜− ⎟ ⎜ ⎟ 

4 

⎟ 

⎝ ⎝ ⎠⎠ 

⎝4 

⎠ 

3 

56. a1 

= 1, r =− 

4 


a1 

1 1 4 

S∞ 

= = = = 

1− r ⎛ ⎛ 3 ⎞⎞ ⎛7 

⎞ 7 

⎜1−⎜− ⎟ ⎜ ⎟ 

4 

⎟ 

⎝ ⎝ ⎠⎠ 

⎝4 

⎠ 

57. a 1 = 8 , 

3 

r = 

2 

Since r > 1, the series diverges. 

58. a 1 = 9 , 

4 

r = 

3 


1 

59. a1 

= 5, r = 

4 


a1 

5 5 20 

S∞ 

= = = = 

1− r ⎛ 1⎞ ⎛3⎞ 

3 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 4 ⎠ ⎝ 4 ⎠ 

1 

60. a1 

= 8, r = 

3 


a1 

8 8 

S∞ 

= = = = 12 

1− r ⎛ 1⎞ ⎛2⎞ 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 3 ⎠ ⎝ 3 ⎠ 

1 

61. a 

1 

= , r = 3 

2 


62. a 1 = 3 , 

3 

r = 

2 






2 

63. a1 

= 6, r = − 

3 


a1 

6 6 18 

S∞ 

= = = = 

1− r ⎛ ⎛ 2 ⎞⎞ ⎛5 

⎞ 5 

⎜1−⎜− ⎟ ⎜ ⎟ 

3 

⎟ 

⎝ ⎝ ⎠⎠ 

⎝3 

⎠ 

1 

64. a1 

= 4, r =− 

2 


a1 

4 4 8 

S∞ 

= = = = 

1− r ⎛ ⎛ 1 ⎞⎞ ⎛3 

⎞ 3 

⎜1−⎜− ⎟ ⎜ ⎟ 

2 

⎟ 

⎝ ⎝ ⎠⎠ 

⎝2 

⎠ 

65. 

66. 

∞ k ∞ k−1 ∞ k−1 

⎛2⎞ 2 ⎛2⎞ ⎛2⎞ 

∑3⎜ ⎟ = ∑3⋅ ⋅ ⎜ ⎟ = ∑ 2⎜ ⎟ 

k= 1 ⎝3⎠ k= 1 3 ⎝3⎠ k= 

1 ⎝3⎠ 

2 

a 1 = 2 , r = 

3 


a1 

2 2 

S∞ 

= = = = 6 

1− 

r 1− 

2 1 

3 3 

∞ k ∞ k−1 ∞ k−1 

⎛3⎞ 3 ⎛3⎞ 3⎛3⎞ 

∑2⎜ ⎟ = ∑2⋅ ⋅ ⎜ ⎟ = ∑ ⎜ ⎟ 

k= 1 ⎝4⎠ k= 1 4 ⎝4⎠ k= 

12⎝4⎠ 

3 3 

a 

1 

= , r = 

2 4 


3 3 

a1 

2 2 3 

S∞ 

= = = = ⋅ 4= 

6 

1− 

r 1− 

3 1 2 4 4 

67. { n + 2 } 

d = ( n+ 1+ 2) − ( n+ 2) = n+ 3−n− 2= 

1 


constant. Therefore the sequence is arithmetic. 

50 50 50 

S = ( k + 2) = k+ 

2 

50 

68. { 2n − 5} 

∑ ∑ ∑ 

k= 1 k= 1 k= 

1 

50(50 + 1) 

= + 2(50) = 1275 + 100 = 1375 

2 

d = 2( n+ 1) −5 −(2n−5) 

= 2n+ 2−5− 2n+ 5= 

2 



50 

50 50 50 

∑ ∑ ∑ 

S = (2k − 5) = 2 k− 

5 

k= 1 k= 1 k= 

1 

⎛50(50 + 1) ⎞ 

= 2⎜ 

⎟− 5(50) = 2550 − 250 = 2300 

⎝ 2 ⎠ 

2 

69. { 4n } Examine the terms of the sequence: 4, 

16, 36, 64, 100, ... 

There is no common difference and there is no 

common ratio. Therefore the sequence is neither 

arithmetic nor geometric. 

2 

70. { 5 1} 

71. 

72. 

n + Examine the terms of the sequence: 

6, 21, 46, 81, 126, ... 




⎧ 2 ⎫ 

⎨ 3 − n ⎬ 

⎩ 3 ⎭ 

⎛ 2 ⎞ ⎛ 2 ⎞ 

d = ⎜3 − ( n+ 1) ⎟−⎜3− 

n⎟ 

⎝ 3 ⎠ ⎝ 3 ⎠ 

2 2 2 2 

= 3− n− − 3+ n =− 

3 3 3 3 



50 50 50 

⎛ 2 ⎞ 2 

S50 

= ∑⎜ 

3− k ⎟= ∑3− 

∑k 

k= 1⎝ 

3 ⎠ k= 1 3 k= 

1 

2 ⎛50(50 + 1) ⎞ 

= 3(50) − ⎜ ⎟= 150 − 850 =−700 

3⎝ 

2 ⎠ 

⎧ 3 ⎫ 

⎨ 8 − n ⎬ 

⎩ 4 ⎭ 

⎛ 3 ⎞ ⎛ 3 ⎞ 

d = ⎜8 − ( n+ 1) ⎟−⎜8− 

n⎟ 

⎝ 4 ⎠ ⎝ 4 ⎠ 

3 3 3 3 

= 8− n− − 8+ n =− 

4 4 4 4 



50 50 50 

S ⎛ 3 3 

50 

8 8 

1 4 k ⎞ 

= ∑⎜ 

− ⎟= ∑ − ∑ 

k k 1 4 

k 

= ⎝ ⎠ = k= 

1 

3 ⎛50(50 + 1) ⎞ 

= 8(50) − ⎜ ⎟ 

4⎝ 

2 ⎠ 

= 400 − 956.25 =−556.25 





73. 1, 3, 6, 10, ... 




74. 2, 4, 6, 8, ... 

The common difference is 2. 



50 50 

⎛50(50 + 1) ⎞ 

S50 

= ∑( 2k ) = 2∑ 

k = 2⎜ 

⎟= 

2550 

k= 1 k= 

1 ⎝ 2 ⎠ 

⎧ 

n 

⎪⎛2 

⎞ ⎫⎪ 

75. ⎨⎜ 

⎟ ⎬ 

⎪⎩⎝3 

⎠ ⎪⎭ 

n+ 

1 

⎛2 

⎞ 

⎜ ⎟ 

n+− 

1 n 

3 ⎛2⎞ 

2 

r = 

⎝ ⎠ 

= 

n ⎜ ⎟ = 

⎛2 

⎞ ⎝3⎠ 

3 

⎜ ⎟ 

⎝3 

⎠ 

The ratio of consecutive terms is constant. 

Therefore the sequence is geometric. 

50 

⎛2 

⎞ 

1 

50 k − 

2 2 ⎜ ⎟ 

⎛ ⎞ 3 

S50 

= 

⎝ ⎠ 

∑ ⎜ ⎟ = ⋅ = 1.999999997 

k = 1 ⎝ 3⎠ 3 2 

1− 

3 

⎧ 

n 

⎪⎛5 

⎞ ⎫⎪ 

76. ⎨⎜ 

⎟ ⎬ 

⎪⎩⎝4 

⎠ ⎪⎭ 

n+ 

1 

⎛5 

⎞ 

⎜ ⎟ 

n+− 

1 n 

4 ⎛5⎞ 

5 

r = 

⎝ ⎠ 

= 

n ⎜ ⎟ = 

⎛5 

⎞ ⎝4⎠ 

4 

⎜ ⎟ 

⎝4 

⎠ 



50 

⎛5 

⎞ 

1 

50 k − 

5 5 ⎜ ⎟ 

⎛ ⎞ 4 

S50 

= 

⎝ ⎠ 

∑ ⎜ ⎟ = ⋅ ≈350,319.6161 

k = 1 ⎝ 4⎠ 4 5 

1− 

4 

77. –1, 2, –4, 8, ... 

2 −4 8 

r = = = =−2 

−1 2 −4 



50 

50 

k −1 

1 −( −2) 

S50 

= ∑ −1( ⋅ − 2) = −1⋅ 

k = 1 

1 − ( − 2) 

14 

≈ 3.752999689× 

10 

78. 1, 1, 2, 3, 5, 8, ... 




/2 

79. { } 

3 n ⎛n+ 

1⎞ 

⎜ ⎟ 

2 

⎛n+ 

1 n⎞ 

3⎝ ⎠ ⎜ − ⎟ 

⎝ 2 2⎠ 

⎛n 

⎞ 

⎜ ⎟ 

⎝2 

⎠ 

r = = 3 = 3 

1/2 

3 



S 

1− 

3 

1/2 

( ) 50 

50 

k /2 1/2 

50 

= ∑ 3 = 3 ⋅ 

1/2 

k = 1 

1− 

3 


80. {( − 1) n 

} 

≈ 2.004706374× 

10 

n+ 

1 

( −1) 

n+− 

1 n 

r = = ( − 1) = −1 

n 

( −1) 



S 

50 

50 

50 

k 1 −( −1) 

= ∑ ( − 1) = ( −1) ⋅ = 0 

1 −( −1) 

k = 1 

81. Find the common ratio of the terms and solve the 

system of equations: 

x+ 2 x+ 

3 

= r; 

= r 

x x+ 

2 

x+ 2 x+ 

3 2 2 

= → x + 4x+ 4= x + 3x→ x = −4 

x x+ 

2 

82. Find the common ratio of the terms and solve the 

system of equations: 

x x+ 

2 

= r; 

= r 

x−1 

x 

x+ 2 x = → 

2 2 

x + x − 2 = x → x = 2 

x x−1 

83. This is a geometric series with 

a1 = $18,000, r = 1.05, n = 5 . Find the 5th 

term: 

− 

( ) ( ) 

5 1 4 

a 5 = 18000 1.05 = 18000 1.05 = $21,879.11 

84. This is a geometric series with 

a1 = $15,000, r = 0.85, n = 6 . Find the 6th 

term: 

− 

( ) ( ) 

6 1 5 

a 6 = 15000 0.85 = 15000 0.85 = $6655.58 





85. a. Find the 10th term of the geometric 

sequence: 

a1 

= 2, r = 0.9, n= 

10 

10−1 9 

a = 2(0.9) = 2(0.9) = 0.775 feet 

10 

b. Find n when a n < 1: 

n−1 

2(0.9) < 1 

n−1 

( 0.9) 

< 0.5 

( ) ( ) 

log ( 0.5) 

n − 1 > 

log ( 0.9) 

log ( 0.5) 

n 

log ( 0.9) 

( n − 1)log 0.9 < log 0.5 

> + 1≈7.58 

On the 8th swing the arc is less than 1 foot. 

c. Find the sum of the first 15 swings: 

15 

⎛ 

15 

1 − (0.9) ⎞ ⎛1− 

( 0.9) 

⎞ 

S15 

= 2 2 

⎜ 1 0.9 ⎟ 

= 

⎝ − ⎠ 

⎜ 0.1 ⎟ 

⎝ ⎠ 

15 

= 20 1− 0.9 = 15.88 feet 

( ( ) ) 

d. Find the infinite sum of the geometric series: 

2 2 

S ∞ = = = 20 feet 

1− 

0.9 0.1 

86. a. Find the 3rd term of the geometric 

sequence: 

a1 

= 24, r = 0.8, n = 3 

3−1 2 

a = 24(0.8) = 24(0.8) = 15.36 feet 

3 

b. The height after the n th bounce is: 

n−1 

−1 

n 

a = 24(0.8) = 24 0.8 0.8 

n 

( ) 

n 

= 30 0.8 ft 

c. Find n when a n < 0.5 : 

n−1 

24(0.8) < 0.5 

( ) ( ) 

n−1 

( 0.8) 

< 0.020833 

( ) ( ) 

log ( 0.020833) 

n − 1 > 

log ( 0.8) 

log ( 0.020833) 

n 

log ( 0.8) 

( n − 1)log 0.8 < log 0.020833 

> + 1≈18.35 

On the 19th bounce the height is less than 

0.5 feet. 

d. Find the infinite sum of the geometric 

series: 

24 24 

S ∞ = = = 120 feet on the upward 

1− 

0.8 0.2 

bounce. 

For the downward motion of the ball: 

30 30 

S ∞ = = = 150 feet 

1− 

0.8 0.2 

The total distance the ball travels is 

120 + 150 = 270 feet. 

87. This is a geometric sequence with 

a1 = 1, r = 2, n = 64 . 

Find the sum of the geometric series: 

⎛ 

64 64 

1−2 ⎞ 1−2 

64 

S64 

= 1 ⎜ 

2 1 

1 2 ⎟ 

= = − 

⎝ − ⎠ −1 

19 

= 1.845× 

10 grains 

88. This is an infinite geometric series with 

a 1 1 

1 = , r = . 

4 4 

Find the sum of the infinite geometric series: 

1 1 

( ) ( ) 1 

S 

4 4 

∞ = = = 

1 3 

( 1− 

) ( ) 

3 

4 4 

Therefore, 1 3 

of the square is eventually shaded. 

89. The common ratio, r = 0.90 < 1. The sum is: 

1 1 

S = = = 10 . 

1− 

0.9 0.10 

The multiplier is 10. 

90. The common ratio, r = 0.95 < 1. The sum is: 

1 1 

S = = = 20 . 

1− 

0.95 0.05 

The multiplier is 20. 


1.03 

a = 4, and r = . 

1.09 

Find the sum: 

4 

Price = $72.67 

1.03 

≈ per share. 

⎛ ⎞ 

⎜1− 

⎟ 

⎝ 1.09 ⎠ 






1.04 

a1 

= 2.5, and r = . 

1.11 

Find the sum: 

2.5 

Price = ≈$39.64 

per share. 

⎛ 1.04 ⎞ 

⎜1− 

⎟ 

⎝ 1.11 ⎠ 

93. Given: a1 = 1000, r = 0.9 

Find n when a n < 0.01 : 

n−1 

1000(0.9) < 0.01 

n−1 

( 0.9) 

< 0.00001 

( ) ( ) 

log ( 0.00001) 

n − 1 > 

log ( 0.9) 

log ( 0.00001) 

n 

log ( 0.9) 

( n − 1)log 0.9 < log 0.00001 

> + 1 ≈110.27 

On the 111th day or December 20, 2007, the 

amount will be less than $0.01. 


111 

⎛ 

n 

1− 

r ⎞ ⎛1− 

( 0.9) 

⎞ 

S111 = a1 

1000⎜ 

⎟ 

⎜ 1 r ⎟ 

= 

⎝ − ⎠ 

⎜ 1−0.9 

⎟ 

⎝ ⎠ 

⎛ 

111 

1− 

( 0.9) 

⎞ 

= 1000 ⎜ ⎟= 

$9999.92 

⎜ 0.1 ⎟ 

⎝ ⎠ 

94. Both options are geometric sequences: 

Option A: a1 = $20,000; r = 1.06; n = 5 

5−1 4 

a5 

= 20,000(1.06) = 20,000(1.06) = $25,250 

⎛ 

5 

1− 

( 1.06) 

⎞ 

S5 

= 20000 ⎜ ⎟= 

$112,742 

⎜ 1−1.06 

⎟ 

⎝ ⎠ 

Option B: a1 = $22,000; r = 1.03; n = 5 

5−1 4 

a5 

= 22,000(1.03) = 22,000(1.03) = $24,761 

⎛ 

5 

1− 

( 1.03) 

⎞ 

S5 

= 22000 ⎜ ⎟= 

$116,801 

⎜ 1−1.03 

⎟ 

⎝ ⎠ 

Option A provides more money in the 5th year, 

while Option B provides the greatest total 

amount of money over the 5 year period. 

95. Find the sum of each sequence: 

A: Arithmetic series with: 

a1 = $1000, d =− 1, n = 1000 

Find the sum of the arithmetic series: 

1000 

S 1000 = ( 1000 + 1 ) = 500(1001) = $500,500 

2 

B: This is a geometric sequence with 

a 1= 1, r = 2, n = 19. 


⎛ 

19 19 

1−2 ⎞ 1−2 

19 

S19 

= 1 ⎜ 

= = 2 − 1 = $524,287 

1 2 ⎟ 

⎝ − ⎠ −1 

B results in more money. 

96. Option 1: 

Total Salary = $2,000,000(7) + $100,000(7) 

= $14,700,000 

Option 2: Geometric series with: 

a1 = $2,000,000, r = 1.045, n = 7 


⎛1− 

( 1.045) 7 ⎞ 

S = 2,000,000 ⎜ ⎟≈$16,038,304 

⎜ 1−1.045 

⎟ 

⎝ ⎠ 

Option 3: Arithmetic series with: 

a1 = $2,000,000, d = $95,000, n = 7 

Find the sum of the arithmetic series: 

7 

S 7 = ( 2(2,000,000) + (7 − 1)(95,000) ) 

2 

= $15,995,000 

Option 2 provides the most money; Option 1 

provides the least money. 

97. The amount paid each day forms a geometric 

sequence with a 1 = 0.01 and r = 2 . 

22 22 

1−r 

1−2 

S22 = a1 

⋅ = 0.01⋅ = 41,943.03 

1− 

r 1− 

2 

The total payment would be $41,943.03 if you 

worked all 22 days. 

( ) 21 

22 1 

a22 = a1 ⋅ r − = 0.01 2 = 20,971.52 

The payment on the 22 nd day is $20,971.52. 

Answers will vary. With this payment plan, the 

bulk of the payment is at the end so missing even 

one day can dramatically reduce the overall 

payment. Notice that with one sick day you 

would lose the amount paid on the 22 nd day 

which is about half the total payment for the 22 

days. 





98. Yes, a sequence can be both arithmetic and 

geometric. For example, the constant sequence 

3,3,3,3,..... can be viewed as an arithmetic 

sequence with a 1 = 3 and d = 0. Alternatively, 

the same sequence can be viewed as a geometric 

sequence with a 1 = 3 and r = 1. 

99 – 100. Answers will vary. 

101. Answers will vary. Both increase (or decrease) 

exponentially, but the domain of a geometric 

sequence is the set of natural numbers while the 

domain of an exponential function is the set of 

all real numbers. 


1. I: n = 1: 2⋅ 1= 2 and1(1 + 1) = 2 

II: If 2+ 4+ 6+ + 2 k = k( k+ 

1) , then 

2+ 4+ 6+ + 2k+ 2( k+ 

1) 

= ( 2+ 4+ 6+ + 2k) 

+ 2( k+ 

1) 

= kk ( + 1) + 2( k+ 

1) 

2 

= k + 3k 

+ 2 

= ( k + 1)( k + 2) 

= ( k+ 1) ⎡⎣( k+ 1) 

+ 1⎤⎦ 

Conditions I and II are satisfied; the statement is 

true. 

2. I: n = 1: 4⋅1− 3= 1and1(2⋅1− 1) = 1 

II: If 1+ 5+ 9 + + (4k − 3) = k(2k−1) 

, then 

1+ 5+ 9 + + (4k− 3) + [4( k+ 1) −3] 

= [ 1+ 5+ 9 + + (4k− 3) ] + 4k+ 4−3 

= k(2k− 1) + 4k 

+ 1 

2 

= 2k − k+ 4k+ 

1 

2 

= 2k 

+ 3k+ 

1 

= ( k+ 1)(2k 

+ 1) 

= ( k+ 1) ⎡⎣2( k + 1) 

−1⎤⎦ 


true. 

3. I: 

1 

n = 1: 1+ 2 = 3and ⋅ 1(1 + 5) = 3 

2 

1 

II: If 3+ 4+ 5 + + ( k+ 2) = ⋅ k( k+ 


2 

3+ 4+ 5 + + ( k+ 2) + [( k+ 1) + 2] 

= [ 3+ 4+ 5 + + ( k+ 2) ] + ( k+ 

3) 

1 

= ⋅ kk ( + 5) + ( k+ 

3) 

2 

1 2 5 

= k + k + k+ 

3 

2 2 

1 2 7 

= k + k + 3 

2 2 

1 2 

= ⋅ ( k + 7k+ 

6) 

2 

= 1 ⋅ ( k+ 1)( k + 6) 

2 

= 1 ⋅ ( k+ 1)[( k + 1) + 5] 

2 


true. 

4. I: n = 1: 2⋅ 1+ 1= 3and1(1 + 2) = 3 

II: If 3+ 5+ 7 + + (2k+ 1) = k( k+ 


3 + 5 + 7 + + (2k+ 1) + [2( k+ 1) + 1] 

= [ 3+ 5+ 7 + + (2k+ 1) ] + (2k+ 

3) 

= kk ( + 2) + (2k+ 

3) 

2 

= k + 2k+ 2k+ 

3 

2 

= k + 4k+ 

3 

= ( k + 1)( k + 3) 

= ( k + 1)[( k+ 1) + 2] 


true. 




Section 12.4: Mathematical Induction 

5. I: 

6. I: 

1 

n = 1: 3⋅1− 1= 2 and ⋅1(3⋅ 1+ 1) = 2 

2 

1 

II: If 2+ 5+ 8 + + (3k− 1) = ⋅ k(3k+ 


2 

2 + 5 + 8 + + (3k− 1) + [3( k+ 1) −1] 

= [ 2+ 5+ 8 + + (3k− 1) ] + (3k+ 

2) 

1 3 2 1 

= ⋅ k(3k+ 1) + (3k+ 2) = k + k+ 3k+ 

2 

2 2 2 

3 2 7 1 2 

= k + k+ 2= ( 3k + 7k+ 

4) 

2 2 2 

= 1 ( k+ 1)(3 k+ 

4) 

2 

1 

= ( k+ 1)[3( k + 1) + 1] 

2 


true. 

1 

n = 1: 3⋅1− 2 = 1and ⋅1(3⋅1− 1) = 1 

2 

1 

II: If 1+ 4+ 7 + + (3k− 2) = ⋅k(3k−1) 

, 

2 

then 

1+ 4+ 7 + + (3k− 2) + [3( k+ 1) −2] 

= [ 1+ 4+ 7 + + (3k− 2) ] + (3k+ 

1) 

1 3 2 1 

= ⋅k(3k − 1) + (3k+ 1) = k − k+ 3k+ 

1 

2 2 2 

3 2 5 1 2 

= k + k + 1= ( 3k + 5k+ 

2) 

2 2 2 

= 1 ( k+ 1)(3 k+ 

2) 

2 

1 

= ( k+ 1)[3( k + 1) − 1] 

2 


true. 

7. I: 

11 − 

1 

n = 1: 2 = 1and 2 − 1= 

1 

2 k−1 

II: If 1+ 2+ 2 + + 2 = 2 −1, then 

2 k− 

1 k+ 1−1 

1+ 2+ 2 + + 2 + 2 

2 k−1 

= ⎡1 2 2 2 ⎤ 

⎣ 

+ + + + 

⎦ 

+ 2 

k k k 

= 2 − 1+ 2 = 2⋅2 −1 

k + 1 

= 2 −1 


true. 

k 

k 

8. I: 

11 − 1 1 

n = 1: 3 = 1and (3 − 1) = 1 

2 

2 k−1 1 

II: If 1+ 3+ 3 + + 3 = ⋅(3 −1) 


2 

2 k− 

1 k+ 1−1 

1+ 3+ 3 + + 3 + 3 

2 k−1 

k 

= ⎡1 3 3 3 ⎤ 

⎣ 

+ + + + 

⎦ 

+ 3 

1 k k 1 k 1 k 

= ⋅(3 − 1) + 3 = ⋅3 − + 3 

2 2 2 

3 k 1 1 k 

= ⋅3 − = ⋅( 3⋅3 −1) 

2 2 2 

1 1 

( 3 

k + 

= − 1 ) 

2 


true. 

11 − 

1 

9. I: n 

1 

( ) 

= 1: 4 = 1and ⋅ 4 − 1 = 1 

3 

2 k−1 1 k 

II: If 1+ 4+ 4 + + 4 = ⋅( 4 −1) 

2 k−1 

( ) 

( ) 

3 

2 k− 

1 k+ 1−1 

1+ 4+ 4 + + 4 + 4 

= 1+ 4+ 4 + + 4 + 4 

1 1 1 

= ⋅ 4 − 1 + 4 = ⋅4 − + 4 

3 3 3 

4 k 1 1 k 

= ⋅4 − = ( 4⋅4 −1) 

3 3 3 

1 1 

( 4 

k + 

= ⋅ − 1 ) 

3 

k k k k 

k 

k 



true. 

11 − 

1 

10. I: n 

1 

( ) 

= 1: 5 = 1and ⋅ 5 − 1 = 1 

4 

2 k−1 1 k 

II: If 1+ 5+ 5 + + 5 = ⋅( 5 −1) 

4 

2 k− 

1 k+ 1−1 

1+ 5+ 5 + + 5 + 5 

2 k−1 

k 

= ⎡1 5 5 5 ⎤ 

⎣ 

+ + + + 

⎦ 

+ 5 

1 1 1 

= ⋅( 5 − 1) 

+ 5 = ⋅5 − + 5 

4 4 4 

5 k 1 1 k 

= ⋅5 − = ( 5⋅5 −1) 

4 4 4 

1 1 

( 5 

k + 

= ⋅ − 1 ) 

4 

k k k k 



true. 





1 1 1 1 

11. I: n = 1: = and = 

1(1+ 1) 2 1+ 

1 2 

1 1 1 1 k 

II: If + + + + = , then 

12 ⋅ 23 ⋅ 34 ⋅ kk ( + 1) k+ 

1 

1 1 1 1 1 

+ + + + + 

1⋅2 2 ⋅3 3⋅ 4 kk ( + 1) ( k+ 1)( k+ 1 + 1) 

⎡ 1 1 1 1 ⎤ 1 

= ⎢ + + + + ⎥+ 

⎣12 ⋅ 23 ⋅ 34 ⋅ kk ( + 1) ⎦ ( k+ 1)( k+ 

2) 

2 2 

k 1 k k + 2 1 k + 2k+ 1 ( k+ 1) k+ 1 k+ 

1 

= + = ⋅ + = = = = 

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 

Conditions I and II are satisfied; the statement is true. 

( ) 

12. I: 

1 1 1 1 

n = 1: = and = 

(2⋅1−1)(2⋅ 1+ 1) 3 2⋅ 1+ 

1 3 

13. I: 

1 1 1 1 k 

II: If + + + + = , then 

1⋅3 3⋅5 5⋅7 (2k− 1)(2k + 1) 2k+ 

1 

1 1 1 1 1 

+ + + + + 

1⋅3 3⋅5 5 ⋅7 (2k − 1)(2k + 1) (2( k+ 1) − 1)(2( k+ 1) + 1) 

⎡ 1 1 1 1 ⎤ 1 

= ⎢ + + + + ⎥+ 

⎣1⋅3 3⋅5 5⋅7 (2k− 1)(2k+ 1) ⎦ (2k+ 1)(2k+ 

3) 

k 1 k 2k+ 

3 1 

= + = ⋅ + 

2k + 1 (2k + 1)(2k+ 3) 2k+ 1 2k+ 3 (2k+ 1)(2k+ 

3) 

2 

2k + 3k+ 1 ( k+ 1)(2k+ 1) k+ 

1 k + 1 

= = = = 

(2k + 1)(2k + 3) (2k+ 1)(2k+ 

3) 2k 

+ 3 2 k+ 1 + 1 


2 1 

n = 1: 1 = 1and ⋅ 1(1 + 1)(2⋅ 1+ 1) = 1 

6 

2 2 2 2 1 

( ) 

II: If 1 + 2 + 3 + + k = ⋅ k( k+ 1)(2k+ 


6 

2 2 2 2 2 2 2 2 2 2 1 

2 

1 + 2 + 3 + + k + ( k+ 1) = ( 1 + 2 + 3 + + k ) + ( k+ 1) = k( k+ 1)(2k+ 1) + ( k+ 

1) 

6 

⎡1 ⎤ ⎡1 2 1 ⎤ ⎡1 2 7 ⎤ 1 

2 

= ( k + 1) ⎢ k(2k+ 1) + k+ 1 ( k 1) k k k 1 ( k 1) k k 1 ( k 1) 2k 7k 

6 

6 ⎥ = + ⎢ + + + 

3 6 ⎥ = + ⎢ + + 

3 6 ⎥ = + + + 

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 6 

1 

= ⋅ ( k + 1)( k+ 2)(2 k+ 

3) 

6 

= 1 ⋅ ( k + 1)[( k+ 1) + 1][2( k+ 1) + 1] 

6 


( ) 





3 1 2 2 

14. I: n = 1: 1 = 1and ⋅ 1 (1+ 1) = 1 

4 

II: If 1 3 + 2 3 + 3 3 + + k 3 = 1 k 2 ( k+ 

1) 

2 , then 

4 

1 

1 + 2 + 3 + + k + ( k+ 1) = ⎡1 2 3 k ⎤ 

⎣ 

+ + + + ⎦ 

+ ( k+ 1) = k ( k+ 1) + ( k+ 

1) 

4 

2 ⎡1 2 ⎤ 1 2 2 

= ( k + 1) k k 1 ( k 1) ⎡k 4k 

4⎤ 

⎢ + + 

4 ⎥ = + + + 

⎣ ⎦ 4 ⎣ ⎦ 

1 ( 1) 

2 ( 2) 

2 

= ⋅ k+ k+ 

4 

1 ( 1) 2 (( 1) 1) 2 

= ⋅ k+ k+ + 

4 

15. I: 

16. I: 

3 3 3 3 3 3 3 3 3 3 2 2 3 


1 

n = 1: 5− 1= 4 and ⋅1(9 − 1) = 4 

2 

1 

II: If 4+ 3+ 2 + + (5 − k) = ⋅k(9 −k) 


2 

1 

4+ 3+ 2 + + (5 − k) + [5 − ( k+ 1)] = [ 4+ 3+ 2 + + (5 − k) ] + (4 − k) = k(9 − k) + (4 −k) 

2 

9 1 2 1 2 7 1 2 

= k − k + 4− k =− k + k+ 4= − ⎡k 7k 

8⎤ 

2 2 2 2 2⎣ 

− − 

⎦ 

1 1 1 

=− ( k+ 1)( k− 8) = ( k+ 1)(8 − k) = ( k+ 1) [ 9 − ( k+ 

1) ] 

2 2 2 


1 

n = 1: − (1 + 1) =−2 and − ⋅ 1(1+ 3) =− 2 

2 

1 

II: If −2−3−4 − 

− ( k+ 1) =− ⋅ k( k+ 


2 

−2−3−4 −− ( k+ 1) − (( k+ 1) + 1) = [ −2−3−4 − 

− ( k+ 1) ] − ( k+ 

2) 

1 1 2 3 1 2 5 

=− ⋅ kk ( + 3) − ( k+ 2) =− k − k−k− 2=− k − k−2 

2 2 2 2 2 

1 2 

1 

=− ⋅ ⎡k 5k 4 ⎤ ( k 1)( k 4) 

2 ⎣ 

+ + 

⎦ 

=− ⋅ + + 

2 

1 

=− ⋅ ( k + 1)[( k+ 1) + 3] 

2 






17. I: 

18. I: 

1 

n = 1: 1(1+ 1) = 2 and ⋅ 1(1 + 1)(1+ 2) = 2 

3 

1 

II: If 12 ⋅ + 23 ⋅ + 34 ⋅ + + kk ( + 1) = ⋅ kk ( + 1)( k+ 


3 

12 ⋅ + 23 ⋅ + 34 ⋅ + + kk ( + 1) + ( k+ 1)( k+ 1+ 1) = [ 12 ⋅ + 23 ⋅ + 34 ⋅ + + kk ( + 1) ] + ( k+ 1)( k+ 

2) 

1 ⎡1 

⎤ 

= ⋅ kk ( + 1)( k+ 2) + ( k+ 1)( k+ 2) = ( k+ 1)( k+ 2) k 1 

3 ⎢ + 

3 ⎥ 

⎣ ⎦ 

1 

= ⋅ ( k + 1)( k+ 2)( k+ 

3) 

3 

1 

= ⋅ ( k + 1)[( k + 1) + 1][( k+ 1) + 2] 

3 


1 

n = 1: (2⋅1−1)(2⋅ 1) = 2and ⋅ 1(1+ 1)(4⋅1− 1) = 2 

3 

1 

II: If 1⋅ 2 + 3⋅ 4 + 5⋅ 6 + + (2k− 1)(2 k) = ⋅ k( k+ 1)(4k−1) 


3 

1⋅ 2 + 3⋅ 4 + 5⋅ 6 + + (2k− 1)(2 k) + [2( k+ 1) − 1][2( k+ 

1)] 

= [ 12 ⋅ + 34 ⋅ + 56 ⋅ + + (2k− 1)(2) k ] + (2k+ 1)( k+ 1)2 ⋅ 

1 ⎡1 

⎤ 

= kk ( + 1)(4k− 1) + 2( k+ 1)(2k+ 1) = ( k+ 1) k(4k 1) 2(2k 

1) 

3 ⎢ − + + 

3 

⎥ 

⎣ 

⎦ 

⎡4 2 1 ⎤ 1 

2 

= ( k + 1) ⎢ k − k+ 4k+ 2 ( k 1) ( 4k k 12k 

6) 

3 3 ⎥ = + − + + 

⎣ 

⎦ 3 

1 ( 1) 

2 

1 

= k + ( 4 k + 11 k+ 

6 ) = ( k+ 1)( k+ 2)(4 k+ 

3) 

3 

3 

1 

= ( k + 1)[( k+ 1) + 1][4( k+ 1) − 1] 

3 


_________________________________________________________________________________________________ 

19. I: 

2 

n = 1: 1 + 1 = 2 is divisible by 2 

20. I: 

3 

n = 1: 1 + 2⋅ 1 = 3 is divisible by 3 

2 

II: If k + k is divisible by 2 , then 

2 2 

( k + 1) + ( k + 1) = k + 2k+ 1+ k+ 

1 

2 

= ( k + k) + (2k+ 

2) 

2 

Since k + k is divisible by 2 and 2k + 2 is 

2 

divisible by 2, then ( k + 1) + ( k + 1) is 

divisible by 2. 


true. 

3 

II: If k + 2k 

is divisible by 3 , then 

3 

( k+ 1) + 2( k+ 

1) 

3 2 

= k + 3k + 3k+ 1+ 2k+ 

2 

3 2 

= ( k + 2 k) + (3k + 3k+ 

3) 

Since k 

2 

3 

+ 2k 

is divisible by 3 and 

3k 

+ 3k+ 3 is divisible by 3, then 

3 

( k + 1) + 2( k+ 1) is divisible by 3. 


true. 





21. I: 

2 

n = 1: 1 − 1+ 2 = 2 is divisible by 2 

2 

II: If k − k + 2 is divisible by 2 , then 

2 2 

( k + 1) − ( k+ 1) + 2= k + 2k+ 1−k− 1+ 

2 

2 

= ( k − k+ 2) + (2 k) 

2 

Since k − k + 2 is divisible by 2 and 2k is 

2 

divisible by 2, then ( k + 1) − ( k + 1) + 2 is 

divisible by 2. 


true. 

22. I: n = 1: 1(1 + 1)(1 + 2) = 6 is divisible by 6 

23. I: 

24. I: 

II: If kk ( + 1)( k+ 2) is divisible by 6 , then 

( k + 1)( k+ 1+ 1)( k+ 1+ 

2) 

= ( k + 1)( k+ 2)( k+ 

3) 

= kk ( + 1)( k+ 2) + 3( k+ 1)( k+ 

2). 

Now, kk ( + 1)( k+ 

2) is divisible by 6; 

and since either k + 1 or k + 2 is even, 

3( k + 1)( k + 2) 

is divisible by 6. 

Thus, ( k + 1)( k+ 2)( k+ 

3) 

= k k+ 1 k+ 2 + 3 k+ 1 k+ 

2 

( )( ) ( )( ) 

is divisible by 6. 


true. 

n= 1: If x > 1then x = x > 1. 

II: Assume, for some natural number k, that if 

k 

x > 1 , then x > 1 . 

k + 1 

Then x > 1, for x > 1, 

k+ 

1 k 

x = x ⋅ x > 1⋅ x = x > 1 

↑ 

k 

( x > 1) 


true. 

n= 1: If 0 < x< 1then 0 < x < 1. 

II: Assume, for some natural number k, that if 

k 

0< x < 1, then 0< x < 1. 

Then, for 0 < x < 1, 

k+ 

1 k 

0< x = x ⋅ x< 1⋅ x = x< 

1 

k + 1 

Thus, 0 < x < 1. 


true. 

1 

1 

25. I: 

1 1 

n= 1: a−bis a factor of a − b = a− 

b. 

II: If a− b is a factor of a − b , show that 

a− b is a factor of a − b . 

k+ 1 k+ 

1 k k 

a − b = a⋅a −b⋅b 

= aa ⋅ −ab ⋅ + ab ⋅ −bb 

⋅ 

k k k 

= a a − b + b ( a−b) 

k 

( ) 

k 

k+ 1 k+ 

1 

k k k k 

Since a− b is a factor of a − b and a− 

b 

is a factor of a− b, then a− b is a factor of 

a 

k+ 1 k+ 

1 

− b . 


true. 

26. I: n = 1: 

21 ⋅+ 1 21 ⋅+ 1 3 3 

a+ bis a factor of a + b = a + b . 

⎡ 3 3 2 2 

a + b = ( a+ b)( a − ab+ 

b ) 

⎤ 

⎣ 

⎦ 

II: 

k 

k 

2k+ 1 2k+ 

1 

If a+ b is a factor of a + b , 

show that a+ b is a factor of 

2( k+ 1) + 1 2( k+ 1) 

+ 1 

a + b . 

2( k+ 1) + 1 2( k+ 1) + 1 2k+ 3 2k+ 

3 

a + b = a + b 

= a ⋅ a + a ⋅b −a ⋅ b + b ⋅b 

2 2k+ 1 2k+ 1 2k+ 

1 2 2 

= a a + b −b ( a −b 

) 

2 2k+ 1 2 2k+ 1 2 2k+ 1 2 2k+ 

1 

( ) 

2k+ 1 2k+ 

1 

Since a+ b is a factor of a + b and 

2 2 

a+ b is a factor of a − b [( a− b)( a+ b) 

] , 

2k+ 3 2k+ 

3 

then a+ b is a factor of a + b . 


true. 

27. I: 1 

n = : ( ) 1 

1+ a = 1+ a ≥ 1+ 1⋅ 

a 

II: Assume that there is an integer k for which 

the inequality holds. We need to show that if 

k 

1+ a ≥ 1+ ka then 

( ) 

k + 1 

( 1+ a) ≥ 1+ ( k+ 1) 

a. 

k+ 

1 

k 

( 1+ a) = ( 1+ a) ( 1+ 

a) 

≥ ( 1+ ka)( 1+ 

a) 

2 

= 1+ ka + a + ka 

= 1+ k+ 1 a+ 

ka 

( ) 

( k ) 

≥ 1+ + 1 a 

Conditions I and II are satisfied, the statement is 

true. 

2 





28. 

n = 1: 

2 

1 − 1+ 41 = 41 is a prime number. 

n = 41: 

2 2 

41 − 41+ 41 = 41 is not a prime number. 

29. II: If 2+ 4+ 6+ + 2k = k + k + 2, then 

2+ 4+ 6+ + 2k+ 2( k+ 

1) 

= 2+ 4+ 6+ + 2k 

+ 2k+ 

2 

I: 

30. I: 

[ ] 

2 

= k + k+ 2+ 2k+ 

2 

2 

= ( k + 2k+ 1) + ( k+ 1) + 2 

2 

= ( k + 1) + ( k+ 1) + 2 

2 

n = 1: 2⋅ 1= 2 and1 + 1+ 2 = 4 ≠ 2 

1 

11 − 

⎛1− 

r ⎞ 

n= 1: ar = a and a ⎜ 

= a 

1− 

r ⎟ 

⎝ ⎠ 

k 

2 k −1 ⎛1− 

r ⎞ 

II: If a+ ar+ ar + + ar = a 

⎜ 

1− 

r ⎟ 

, 

⎝ ⎠ 


2 k− 

1 k+ 1−1 

a+ ar+ ar + + ar + ar 

2 k−1 

k 

= ⎡a ar ar ar ⎤ 

⎣ 

+ + + + 

⎦ 

+ ar 

⎛ 

k k k 

1 −r ⎞ 

k a(1 − r ) + ar (1 −r) 

= a 

⎜ 

ar 

1 r ⎟ 

+ = 

⎝ − ⎠ 

1−r 

k k k+ 1 k+ 

1 

a− ar + ar −ar ⎛1−r 

⎞ 

= = a 

1−r 

⎜ 1−r 

⎟ 

⎝ ⎠ 


true. 

31. I: n = 1: 

1(1 −1) 

a+ (1− 1) d = a and 1⋅ a+ d = a 

2 

II: If a+ ( a+ d) + ( a+ 2 d) + + [ a+ ( k−1) 

d] 

kk ( −1) 

= ka + d 

2 


a+ ( a+ d) + ( a+ 2 d) + + [ a+ ( k − 1) d] + ( a+ 

kd) 

= [ a+ ( a+ d) + ( a+ 2 d) + + [ a+ ( k − 1) d] 

] + ( a+ 

kd) 

kk ( −1) 

= ka + d + ( a + kd) 

2 

⎡kk 

( −1) 

⎤ 

= ( k+ 1) a+ d ⎢ + k 

2 ⎥ 

⎣ ⎦ 

2 

⎡ 

2 

k − k + 2k⎤ 

= ( k+ 1) a+ 

d 

⎢⎣ 

2 ⎥⎦ 

⎡ 

2 

k + k⎤ 

= ( k+ 1) a+ 

d 

⎢⎣ 

2 ⎥⎦ 

⎡( k + 1) k⎤ 

= ( k+ 1) a+ 

d⎢ 

2 

⎥ 

⎣ ⎦ 

⎡( k + 1) ⎡ ( k + 1) 

−1⎤⎤ 

= ( k + 1) 

a+ 

d 

⎣ ⎦ 

⎢ 

⎥ 

⎢⎣ 

2 ⎥⎦ 


true. 

32. I: n = 4: The number of diagonals of a 

quadrilateral is 1 4(4 3) 2 

2 ⋅ − = . 

II: Assume that for any integer k, the number of 

diagonals of a convex polygon with k sides 

(k vertices) is 1 ( 3) 

2 ⋅ kk− 

. A convex 

polygon with k + 1 sides ( k + 1 vertices) 

consists of a convex polygon with k sides 

(k vertices) plus a triangle, for a total of 

( k + 1) vertices. The diagonals of this 

k + 1-sided convex polygon consist of the 

diagonals of the k-sided polygon plus k − 1 

additional diagonals. For example, consider 

the following diagrams. 

k + 1 = 6 sides 

k = 5 sides k − 1 = 4 new diagonals 

Thus, we have the equation: 

1 1 2 3 

kk ( − 3) + ( k− 1) = k − k+ k−1 

2 2 2 

1 2 1 

= k − k−1 

2 2 

1 2 

= ( k −k−2) 

2 

1 

= ( k + 1)( k− 

2) 

2 

1 

= ( k + 1)[( k+ 1) − 3] 

2 

Conditions I and II are satisfied; the 

statement is true. 




Section 12.5: The Binomial Theorem 

33. I: n = 3: (3 −2) ⋅ 180°= 180 ° which is the 

sum of the angles of a triangle. 

II: Assume that for any integer k, the sum of 

the angles of a convex polygon with k sides 

is ( k −2) ⋅ 180°. A convex polygon with 

k + 1 sides consists of a convex polygon 

with k sides plus a triangle. Thus, the sum of 

the angles is 

( k −2) ⋅ 180°+ 180 °= [( k+ 1) −2] ⋅ 180 ° . 


true. 



1. Pascal Triangle 

2. 

⎛6⎞ 6! 654321 ⋅ ⋅ ⋅ ⋅ ⋅ 65 ⋅ 

⎜ ⎟= = = = 15 

⎝2⎠ 

2!4! 2⋅1⋅4⋅3⋅2⋅1 2⋅1 

⎛n⎞ n! 

3. False; ⎜ ⎟ = 

⎝ j ⎠ j! ( n− 

j) 

! 

4. Binomial Theorem 

5. 

6. 

⎛5⎞ 5! 54321 ⋅ ⋅ ⋅ ⋅ 54 ⋅ 

⎜ ⎟= = = = 10 

⎝3⎠ 

3! 2! 3⋅2⋅1⋅2⋅1 2⋅1 

⎛7⎞ 7! 7654321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 765 ⋅ ⋅ 

⎜ ⎟= = = = 35 

⎝3⎠ 

3!4! 3214321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 321 ⋅ ⋅ 

7. 

8. 

9. 

10. 

11. 


13. 

14. 

15. 

16. 

⎛7⎞ 7! 7⋅6⋅5⋅4⋅3⋅2⋅1 7⋅6 

⎜ ⎟= = = = 21 

⎝5⎠ 

5!2! 5432121 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 21 ⋅ 

⎛9⎞ 9! 9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 9⋅8 

⎜ ⎟= = = = 36 

⎝7⎠ 

7!2! 765432121 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 21 ⋅ 

⎛50⎞ 50! 50⋅ 

49! 50 

⎜ ⎟= = = = 50 

⎝49⎠ 

49!1! 49! ⋅1 1 

⎛100⎞ 100! 100⋅99⋅98! 100⋅99 

⎜ ⎟= = = = 4950 

⎝ 98 ⎠ 98!2! 98!21 ⋅ ⋅ 21 ⋅ 

⎛1000⎞ 1000! 1 

⎜ ⎟ = = = 1 

⎝1000⎠ 

1000!0! 1 

⎛1000⎞ 1000! 1 

⎜ ⎟ = = = 1 

⎝ 0 ⎠ 0!1000! 1 

⎛55⎞ 55! 

⎜ ⎟ = ≈ 1.8664 × 10 

⎝23⎠ 

23!32! 

⎛60⎞ 60! 

⎜ ⎟ = ≈ 4.1918 × 10 

⎝20⎠ 

20! 40! 

⎛47⎞ 47! 

⎜ ⎟ = ≈ 1.4834 × 10 

⎝25⎠ 

25! 22! 

⎛37⎞ 37! 

⎜ ⎟ = ≈ 1.7673 × 10 

⎝19 ⎠ 19!18! 

15 

15 

13 

10 

_________________________________________________________________________________________________ 

17. 

18. 

19. 

⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ 

( x + 1) = ⎜ ⎟x + ⎜ ⎟x + ⎜ ⎟x + ⎜ ⎟x + ⎜ ⎟x + ⎜ ⎟x 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ 

5 5 4 3 2 1 0 

5 4 3 2 

= x + 5x + 10x + 10x + 5x+ 

1 

⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ 

( x − 1) = ⎜ ⎟x + ⎜ ⎟( − 1) x + ⎜ ⎟( − 1) x + ⎜ ⎟( − 1) x + ⎜ ⎟( − 1) x + ⎜ ⎟( −1) 

x 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ 

5 5 4 2 3 3 2 4 1 5 0 

5 4 3 2 

= x − 5x + 10x − 10x + 5x−1 

⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ 

( x− 2) = ⎜ ⎟x + ⎜ ⎟x ( − 2) + ⎜ ⎟x ( − 2) + ⎜ ⎟x ( − 2) + ⎜ ⎟x ( − 2) + ⎜ ⎟x( − 2) + ⎜ ⎟x 

( −2) 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ ⎝6⎠ 

6 6 5 4 2 3 3 2 4 5 0 6 

6 5 4 3 2 

= x + 6 x ( − 2) + 15x ⋅ 4 + 20 x ( − 8) + 15x ⋅ 16 + 6 x⋅( − 32) + 64 

6 5 4 3 2 

= x − 12x + 60x − 160x + 240x − 192x+ 

64 





20. 

21. 

⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ 

( x+ 3) = ⎜ ⎟x + ⎜ ⎟x (3) + ⎜ ⎟x (3) + ⎜ ⎟x (3) + ⎜ ⎟x (3) + ⎜ ⎟x 

(3) 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ 

5 5 4 3 2 2 3 1 4 0 5 

5 4 3 2 

= x + 5 x (3) + 10x ⋅ 9 + 10 x (27) + 5x⋅ 81+ 

243 

5 4 3 2 

= x + 15x + 90x + 270x + 405x+ 

243 

4 ⎛4⎞ 4 ⎛4⎞ 3 ⎛4⎞ 2 ⎛4⎞ ⎛4⎞ 

(3x+ 1) = ⎜ ⎟(3 x) + ⎜ ⎟(3 x) + ⎜ ⎟(3 x) + ⎜ ⎟(3 x) 

+ ⎜ ⎟ 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ 

4 3 2 4 3 2 

= 81x + 4⋅ 27x + 6⋅ 9x + 4⋅ 3x+ 1 = 81x + 108x + 54x + 12x+ 

1 

22. 

⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ 

(2x+ 3) = ⎜ ⎟(2 x) + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟⋅2x⋅ 3 + ⎜ ⎟⋅3 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ 

5 5 4 3 2 2 3 4 5 

5 4 3 2 

= 32x + 5⋅16x ⋅ 3 + 10⋅8x ⋅ 9 + 10⋅4x ⋅ 27 + 5⋅2x⋅ 81+ 

243 

5 4 3 2 

= 32x + 240x + 720x + 1080x + 810x+ 

243 

2 2 ⎛5⎞ 2 ⎛5⎞ 2 2 ⎛5⎞ 2 2 ⎛5⎞ 2 2 ⎛5⎞ 2 2 ⎛5⎞ 

2 

23. ( x + y ) = ⎜ ⎟( x ) + ⎜ ⎟( x ) y + ⎜ ⎟( x ) ( y ) + ⎜ ⎟( x ) ( y ) + ⎜ ⎟x ( y ) + ⎜ ⎟( y ) 

5 5 4 3 2 2 3 4 5 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ 

10 8 2 6 4 4 6 2 8 10 

= x + 5x y + 10x y + 10x y + 5x y + y 

2 2 ⎛6⎞ 2 ⎛6⎞ 2 2 ⎛6⎞ 2 2 ⎛6⎞ 2 2 ⎛6⎞ 

2 2 

24. ( x − y ) = ⎜ ⎟( x ) + ⎜ ⎟( x ) ( − y ) + ⎜ ⎟( x ) ( − y ) + ⎜ ⎟( x ) ( − y ) + ⎜ ⎟( x ) ( −y 

) 

6 6 5 4 2 3 3 2 4 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ 

⎛6⎞ 

2 2 

5 ⎛6⎞ 

2 

6 

+ ⎜ ⎟x ( − y ) + ⎜ ⎟( −y 

) 

⎝5⎠ 

⎝6⎠ 

12 10 2 8 4 6 6 4 8 2 10 12 

= x − 6x y + 15x y − 20x y + 15x y − 6x y + y 

⎛6⎞ ⎛6⎞ ⎛6⎞ ⎛6⎞ 

25. ( x + 2) = ⎜ ⎟( x) + ⎜ ⎟( x) ( 2) + ⎜ ⎟( x) ( 2) + ⎜ ⎟( x) ( 2) 

6 6 5 1 4 2 3 3 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ 

⎛6⎞ 2 4 ⎛6⎞ 5 ⎛6⎞ 

6 

+ ⎜ ⎟( x) ( 2) + ⎜ ⎟( x)( 2) + ⎜ ⎟( 2) 

⎝4⎠ ⎝5⎠ ⎝6⎠ 

3 5/2 2 3/2 1/2 

= x + 6 2x + 15⋅ 2x + 20⋅ 2 2x + 15⋅ 4x+ 6⋅ 4 2x 

+ 8 

3 5/2 2 3/2 1/2 

= x + 6 2x + 30x + 40 2x + 60x+ 24 2x 

+ 8 

⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ 

26. ( x − 3) = ⎜ ⎟( x) + ⎜ ⎟( x) ( − 3) + ⎜ ⎟( x) ( − 3) + ⎜ ⎟( x)( − 3) + ⎜ ⎟( − 3) 

4 4 3 1 2 2 3 4 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ 

2 3/2 1/2 

= x − 4 3x + 63 ⋅ x−43 ⋅ 3x 

+ 9 

2 3/2 1/2 

= x − 4 3x + 18x− 12 3x 

+ 9 

⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ 

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ 

5 5 4 3 2 2 3 4 5 

27. ( ax + by) = ( ax) + ( ax) ⋅ by + ( ax) ( by) + ( ax) ( by) + ax( by) + ( by) 

5 5 4 4 3 3 2 2 2 2 3 3 4 4 5 5 

= a x + 5a x by+ 10a x b y + 10a x b y + 5axb y + b y 

⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ 

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ 

4 4 3 2 2 3 4 

28. ( ax − by) = ( ax) + ( ax) ( − by) 

+ ( ax) ( − by) + ( ax)( − by) + ( −by) 

4 4 3 3 2 2 2 2 3 3 4 4 

= a x − 4a x by + 6a x b y − 4axb y + b y 




Section 12.5: The Binomial Theorem 

29. n = 10, j = 4, x = x, a = 3 

⎛10⎞ 10! 10987 ⋅ ⋅ ⋅ 

⎜ ⎟x ⋅ 3 = ⋅ 81x = ⋅81x 

⎝ 4 ⎠ 4!6! 4⋅3⋅2⋅1 

6 

= 17,010x 

6 

The coefficient of x is 17,010. 

6 4 6 6 

30. n= 10, j = 7, x = x, a = − 3 

⎛10⎞ 

3 7 10! 

3 

⎜ ⎟x 

⋅− ( 3) = ⋅− ( 2187) 

x 

⎝ 7 ⎠ 7!3! 

10⋅9⋅8 

= ⋅( −2187) 

x 

321 ⋅ ⋅ 

3 

=−262,440x 

3 

The coefficient of x is − 262,440. 

31. n = 12, j = 5, x = 2 x, a = − 1 

⎛12⎞ 

7 5 12! 7 

⎜ ⎟(2 x) ⋅− ( 1) = ⋅128 x ( −1) 

⎝ 5 ⎠ 

5!7! 

12⋅11⋅10⋅9⋅8 = ⋅ ( − 128) x 

54321 ⋅ ⋅ ⋅ ⋅ 

7 

=−101,376x 

7 


32. n= 12, j = 9, x = 2 x, a = 1 

⎛12⎞ 12! 12⋅11⋅10 

⎜ ⎟(2 x) ⋅ (1) = ⋅ 8 x (1) = ⋅8x 

⎝ 9 ⎠ 

9!3! 3⋅2⋅1 

3 

= 1760x 

3 

The coefficient of x is 1760. 

3 9 3 3 

33. n= 9, j = 2, x = 2 x, a = 3 

⎛9⎞ 

7 2 9! 7 

⎜ ⎟(2 x) ⋅ 3 = ⋅128 x (9) 

⎝2⎠ 

2!7! 

98 ⋅ 128 

7 

= ⋅ x ⋅ 9 

21 ⋅ 

7 

= 41,472x 

7 

The coefficient of x is 41,472. 

34. n= 9, j = 7, x = 2 x, a = − 3 

⎛9⎞ 

2 7 9! 2 

⎜ ⎟(2 x) ⋅− ( 3) = ⋅4 x ( −2187) 

⎝7⎠ 

7!2! 

98 ⋅ 4 

2 

= ⋅ x ⋅− 2187 

21 ⋅ 

2 

=−314,928x 

2 


3 

7 

35. n= 7, j = 4, x = x, a = 3 

⎛7⎞ 

7! 7⋅6⋅5 

⎜ ⎟x ⋅ 3 = ⋅ 81x = ⋅ 81x = 2835x 

⎝4⎠ 

4!3! 3⋅2⋅1 

3 4 3 3 3 

36. n= 7, j = 2, x = x, a = − 3 

⎛7⎞ 

7! 7⋅6 

⎜ ⎟x ⋅− ( 3) = ⋅ 9x = ⋅ 9x = 189x 

⎝2⎠ 

2!5! 2⋅1 

5 2 5 5 5 

37. n= 9, j = 2, x = 3 x, a =− 2 

⎛9⎞ 

7 2 9! 

7 

⎜ ⎟(3 x) ⋅− ( 2) = ⋅2187x 

⋅4 

⎝2⎠ 

2!7! 

98 ⋅ 

= ⋅ 8748 x = 314,928 x 

21 ⋅ 

38. n= 8, j = 5, x = 3 x, a = 2 

7 7 

⎛8⎞ 

3 5 8! 3 

⎜ ⎟(3 x) ⋅ ( 2) = ⋅27x 

⋅32 

⎝5⎠ 

5!3! 

876 ⋅ ⋅ 

= ⋅ 864 x = 48,384 x 

321 ⋅ ⋅ 

0 

39. The x term in 

∑ 

12 12 

⎛12⎞ 12− 

j ⎛1 ⎞ 12 

( ) 

j ⎛ ⎞ 

⎜ ⎟ x 

x 

j= 0 

j 

⎜ ⎟ = ∑ ⎜ ⎟ 

⎝ ⎠ ⎝ x ⎠ j= 

0⎝ j ⎠ 

occurs when: 

24 − 3 j = 0 

24 = 3 j 

j = 8 

3 3 

2 24−3j 

⎛12⎞ 12! 12⋅11⋅10 ⋅9 

The coefficient is ⎜ 495 

8 ⎟ = = = 

⎝ ⎠ 8!4! 4321 ⋅ ⋅ ⋅ 

0 


∑ 

9 j 9 

⎛9⎞ − ⎛ 1 ⎞ ⎛9⎞ 

⎜ ⎟ x 

1 

2 

j= 0 

j 

⎜− ⎟ = ∑ ⎜ ⎟ − 

⎝ ⎠ ⎝ x ⎠ j= 

0⎝ j⎠ 

occurs when: 

9− 3j 

= 0 

9= 

3j 

j = 3 

The coefficient is 

⎛9⎞ 

( ) 3 9! 9⋅8⋅7 

⎜ ⎟ − 1 =− =− =−84 

⎝3⎠ 

3!6! 3⋅2⋅1 

( ) ( ) 

9 j j 9−3j 

x 





4 

2 



10 j 

∑ 10 3 

⎛10⎞ 10− 

j ⎛ − 2 ⎞ ⎛10⎞ 

j 10− 

j 

( x) ( 2) 

x 2 

⎜ ⎟ ⎜ ⎟ = ∑ 

8 j 

⎜ ⎟ − 

∑ ⎛8⎞ 8− 

j ⎛ 3 ⎞ 

( ) 8 ⎛8⎞ 

j 4− 

j 

⎜ ⎟ x ⎜ ⎟ = ∑ ⎜ ⎟( 3) 

x 

j= 0⎝ j ⎠ ⎝ x ⎠ j= 

0⎝ j ⎠ 

j= 0⎝ j⎠ ⎝ x ⎠ j= 

0⎝ j⎠ 

occurs when: 

occurs when: 

3 

4− j = 2 

10 − j = 4 

2 

− j =−2 

3 

− j =−6 

j = 2 

2 


j = 4 

⎛8⎞ 


( ) 2 8! 8⋅7 

⎜ ⎟ 3 = ⋅ 9= ⋅ 9= 

252 

⎝2⎠ 

6!2! 2⋅1 

⎛10⎞ ( ) 4 10! 10987 ⋅ ⋅ ⋅ 

⎜ ⎟ − 2 = ⋅ 16 = ⋅ 16 = 3360 

⎝ 4 ⎠ 6!4! 4⋅3⋅2⋅1 

_________________________________________________________________________________________________ 

5 −3 ⎛5⎞ 5 ⎛5⎞ 4 −3 ⎛5⎞ 3 −3 ⎛5⎞ 

2 −3 

43. (1.001) ( 1 10 ) ⎜ ⎟1 ⎜ ⎟1 10 ⎜ ⎟1 ( 10 ) ⎜ ⎟1 ( 10 ) 

5 2 3 

= + = + ⋅ + ⋅ + ⋅ +⋅⋅⋅ 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ 

= 1+ 5(0.001) + 10(0.000001) + 10(0.000000001) +⋅⋅⋅ 

= 1+ 0.005 + 0.000010 + 0.000000010 +⋅⋅⋅ 

= 1.00501 (correct to 5 decimal places) 

6 ⎛6⎞ ⎛6⎞ ⎛6⎞ 2 ⎛6⎞ 

3 

= − = ⎜ ⎟ + ⎜ ⎟ ⋅ − + ⎜ ⎟ ⋅ − + ⎜ ⎟⋅ ⋅ − +⋅⋅⋅ 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ 

= 1+ 6( − 0.002) + 15( 0.000004) + 20( − 0.000000008 ) + ... 

= 1− 0.012 + 0.00006 −0.00000016 

= 0.98806 (correct to 5 decimal places) 

6 6 5 4 3 

44. (0.998) ( 1 0.002) 1 1 ( 0.002) 1 ( 0.002) 1 ( 0.002) 

45. 

( − ) 

( ) 

⎛ n ⎞ n! n! 

n n 1! 

⎜ ⎟ = = = = n ; 

⎝n −1⎠ 

( n−1!( ) n−( n−1)! ) ( n−1!(1)! ) n−1! 

⎛n⎞ n! n! n! n! 

⎜ ⎟ = = = = = 1 

⎝n⎠ 

n!( n−n)! n! 0! n! ⋅1 n! 

46. 

⎛ n ⎞ n! n! n! 

⎛n⎞ 

⎜ ⎟= = = = ⎜ ⎟ 

⎝n− j⎠ ( n− j)!( n−( n− j))! ( n− j) 

! j! j!( n− 

j)! 

⎝ j⎠ 

⎛n⎞ ⎛n⎞ ⎛n⎞ 

n 

47. Show that ⎜ ⎟+ ⎜ ⎟+⋅⋅⋅+ ⎜ ⎟= 

2 

⎝0⎠ ⎝1⎠ ⎝n⎠ 

n n ⎛n⎞ n ⎛n⎞ n−1 ⎛n⎞ n−2 2 ⎛n⎞ n−n n ⎛n⎞ ⎛n⎞ ⎛n⎞ 

2 = (1+ 1) = ⎜ ⎟⋅ 1 + ⎜ ⎟⋅1 ⋅ 1+ ⎜ ⎟⋅1 ⋅ 1 +⋅⋅⋅+ ⎜ ⎟⋅1 ⋅ 1 = ⎜ ⎟+ ⎜ ⎟+⋅⋅⋅+ 

⎜ ⎟ 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝n⎠ ⎝0⎠ ⎝1⎠ ⎝n⎠ 

⎛n⎞ ⎛n⎞ ⎛n⎞ n ⎛n⎞ 

48. Show that ⎜ ⎟− ⎜ ⎟+ ⎜ ⎟−⋅⋅⋅+ ( − 1) ⎜ ⎟= 

0 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝n⎠ 

n ⎛n⎞ n ⎛n⎞ n−1 ⎛n⎞ n−2 2 ⎛n⎞ n−n n ⎛n⎞ ⎛n⎞ ⎛n⎞ n⎛n⎞ 

0 = (1 − 1) = ⎜ ⎟⋅ 1 + ⎜ ⎟⋅1 ⋅− ( 1) + ⎜ ⎟⋅1 ⋅− ( 1) +⋅⋅⋅+ ⎜ ⎟⋅1 ⋅− ( 1) = ⎜ ⎟− ⎜ ⎟+ ⎜ ⎟−⋅⋅⋅+− 

( 1) ⎜ ⎟ 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝n⎠ ⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝n⎠ 

49. 

5 4 3 2 2 3 4 5 5 

⎛5⎞⎛1⎞ ⎛5⎞⎛1⎞ ⎛3⎞ ⎛5⎞⎛1⎞ ⎛3⎞ ⎛5⎞⎛1⎞ ⎛3⎞ ⎛5⎞⎛1⎞⎛3⎞ ⎛5⎞⎛3⎞ ⎛1 3⎞ 

5 

⎜ ⎟ (1) 1 

0 

⎜ ⎟ + ⎜ ⎟ 

4 1 

⎜ ⎟ ⎜ ⎟+ ⎜ ⎟ + + + = + = = 

4 4 2 

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 

4 4 3 

⎜ ⎟ ⎜ ⎟ 

4 4 4 

⎜ ⎟⎜ ⎟ 

4 4 5 

⎜ ⎟ ⎜ ⎟ 

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝4⎠ 

⎝4 4⎠ 




Chapter 12 Review Exercises 

50. 

8 

12! = 479,001,600 = 4.790016× 

10 

18 

20! ≈ 2.432902008× 

10 

25 

25! ≈ 1.551121004× 

10 

⎛12 ⎞ ⎛ 1 ⎞ 

12! ≈ 2⋅12π ⎜ ⎟ ⎜1+ 

⎟ 

⎝ e ⎠ ⎝ 12 ⋅ 12 − 1 ⎠ 

≈ 479,013,972.4 

⎛20 ⎞ ⎛ 1 ⎞ 

20! ≈ 2⋅20π ⎜ ⎟ ⎜1+ 

⎟ 

⎝ e ⎠ ⎝ 12 ⋅ 20 − 1 ⎠ 

18 

≈ 2.43292403× 

10 

⎛25 ⎞ ⎛ 1 ⎞ 

25! ≈ 2⋅25π ⎜ ⎟ ⎜1+ 

⎟ 

⎝ e ⎠ ⎝ 12 ⋅ 25 − 1 ⎠ 

25 

≈ 1.551129917× 

10 


1. 

2. 

3. 

4. 


20 

25 

11+ 3 4 2 2+ 

3 5 

a1 = ( − 1) =− , a2 

= ( − 1) = , 

1+ 2 3 2+ 

2 4 

3 3+ 3 6 4 4+ 

3 7 

a3 = ( − 1) =− , a4 

= ( − 1) = , 

3+ 2 5 4+ 

2 6 

5 5+ 

3 8 

a5 

= ( − 1) =− 5 + 2 7 

11 + 

b1 

= ( −1) (2 ⋅ 1+ 3) = 5, 

2+ 

1 

b2 

= ( −1) (2⋅ 2+ 3) =−7, 

3+ 

1 

b3 

= ( −1) (2⋅ 3+ 3) = 9, 

4+ 

1 

b4 

= ( −1) (2⋅ 4+ 3) =−11, 

5+ 

1 

b = ( −1) (2⋅ 5+ 3) = 13 

5 

1 2 3 

1 2 2 2 3 2 

2 2 2 4 2 8 

c = = = 2, c = = = 1, c = = , 

1 1 2 4 3 9 

4 5 

2 16 2 32 

c4 = = = 1, c 

2 5 = = 

2 

4 16 5 25 

1 2 3 4 5 

1 = e = , 2 = e , 3 = e , e 4 = , 

e 

5 = 

d e d d d d 

1 2 3 4 5 

2 2 4 

a = 3, a = ⋅ 3 = 2, a = ⋅ 2 = , 

3 3 3 

2 4 8 2 8 16 

a4 = ⋅ = , a5 

= ⋅ = 

3 3 9 3 9 27 

5. 1 2 3 

1 1 1 

= 4, =− ⋅ 4 =− 1, =− ⋅− 1 = , 

4 4 4 

1 1 1 1 1 1 

a4 =− ⋅ =− , a5 

=− ⋅− = 

4 4 16 4 16 64 

6. a1 a2 a3 

a = 2, a = 2 − 2 = 0, a = 2 − 0 = 2, 

7. 1 2 3 

a 

= 2− 2= 0, a = 2− 0= 

2 

4 5 

8. a1 a2 a3 

9. 

a 

= − 3, = 4 + ( − 3) = 1, = 4 + 1 = 5, 

= 4+ 5= 9, a = 4+ 9= 

13 

4 5 

4 

∑ 

k = 1 

(4k 

+ 2) 

( 4 1 2) ( 4 2 2) ( 4 3 2) ( 4 4 2) 

( 6) ( 10) ( 14) ( 18) 

= ⋅ + + ⋅ + + ⋅ + + ⋅ + 

= + + + 

= 48 

3 

2 2 2 2 

10. ∑ (3 − k ) = ( 3 − 1 ) + ( 3 − 2 ) + ( 3 −3 

) 

k = 1 

( 2) ( 1) ( 6) 

= + − + − 

=−5 

1 1 1 1 k + 1 ⎛1⎞ 

− + − +⋅⋅⋅+ = − ⎜ ⎟ 

2 3 4 13 

⎝k 

⎠ 

11. 1 ∑ ( 1) 

13 

k = 1 

2 3 4 n+ 1 n 1 

2 2 2 2 ⎛ 

k+ 

2 ⎞ 

12. 2 + + + +⋅⋅⋅+ = 

2 3 

n ∑ 

k 

3 3 3 3 ⎜ 

k = 0 3 ⎟ 

⎝ ⎠ 

n+ 

1⎛ 

k 

2 ⎞ 

= ∑ 

⎜ k −1 

k = 1 3 ⎟ 

⎝ ⎠ 

13. { a } { n 5} 

n 

= + Arithmetic 

d = ( n+ 1+ 5) − ( n+ 5) = n+ 6−n− 5= 

1 

n 

n 

Sn 

= [ 6+ n+ 5 ] = ( n+ 

11) 

2 2 

14. { b } { 4n 

3} 

n 

= + Arithmetic 

d = (4( n+ 1) + 3) − (4n+ 

3) 

= 4n+ 4+ 3−4n− 3= 

4 

n 

Sn 

= + + 

2 

n 

= (4n 

+ 10) 

2 

= n + 

[ 7 4n 

3] 

( 2n 

5) 





3 

15. { cn} { 2n 

} 

= Examine the terms of the 

sequence: 2, 16, 54, 128, 250, ... 

There is no common difference; there is no 

common ratio; neither. 

2 

16. { dn} { 2n 

1} 

= − Examine the terms of the 

sequence: 1, 7, 17, 31, 49, ... 

There is no common difference; there is no 

common ratio; neither. 

3 

17. { } { 2 n 

n } 

s = Geometric 

3( n+ 1) 3n+ 

3 

2 2 3n+ 3−3n 

3 

r = = = 2 = 2 = 8 

3n 

3n 

2 2 

⎛ 

n 

n 

1−8 ⎞ ⎛1−8 ⎞ 8 n 

S n = 8 ⎜ = 8 = 8 −1 

1 8 ⎟ ⎜ 7 ⎟ 

⎝ − ⎠ ⎝ − ⎠ 7 

2 

18. { } { 3 n 

n } 

u = Geometric 

2( n+ 1) 2n+ 

2 

( ) 

3 3 2n+ 2−2n 

2 

r = = = 3 = 3 = 9 

2n 

2n 

3 3 

⎛ 

n 

n 

1−9 ⎞ ⎛1−9 ⎞ 9 n 

S n = 9 ⎜ = 9 = 9 −1 

1 9 ⎟ ⎜ 8 ⎟ 

⎝ − ⎠ ⎝ − ⎠ 8 

( ) 

19. 0, 4, 8, 12, ... Arithmetic d = 4− 0= 

4 

n 

n 

Sn 

= ( 2(0) + ( n− 1)4 ) = ( 4( n− 1) ) = 2 n( n− 

1) 

2 2 

20. 1, –3, –7, –11, ... Arithmetic 

d =−3− 1=− 

4 

21. 

n 

n 

Sn 

= + n− − = 

2 2 

− n+ 

n 

= (6 − 4 n) = n( 3−2n) 

2 

( 2(1) ( 1)( 4) ) ( 2 4 4) 

3 3 3 3 

3, , , , , ... 

2 4 8 16 

⎛3 

⎞ 

⎜ ⎟ 

2 3 1 1 

r = 

⎝ ⎠ 

= ⋅ = 

3 2 3 2 

S n 

Geometric 

⎛ 

n 

n 

⎛1⎞ ⎞ ⎛ ⎛1⎞ 

⎞ 

⎜1−⎜ ⎟ ⎟ ⎜1−⎜ ⎟ ⎟ 

n 

2 2 ⎛ ⎛1 

⎞ ⎞ 

= 3⎜ ⎝ ⎠ ⎟= 3⎜ ⎝ ⎠ ⎟= 6 1− 

⎜ 1 ⎟ ⎜ 1 

⎜ ⎟ 

1 

⎛ ⎞ ⎟ ⎜ ⎝2 

⎠ ⎟ 

− 

⎝ ⎠ 

⎜ 2 ⎟ ⎜ ⎜ ⎟ 

⎝ 2 ⎠ ⎟ 

⎝ ⎠ ⎝ ⎠ 

22. 

5 5 5 5 

5, − , , − , , ... Geometric 

3 9 27 81 

⎛ 5 ⎞ 

⎜− 

⎟ 

3 5 1 1 

r = 

⎝ ⎠ 

=− ⋅ =− 

5 3 5 3 

S 

n 

⎛ 

n 

n 

⎛ 1⎞ ⎞ ⎛ ⎛ 1⎞ 

⎞ 

⎜1−⎜− 

⎟ ⎟ ⎜1−⎜− 

⎟ ⎟ 

3 3 

= 5⎜ ⎝ ⎠ ⎟= 

5⎜ ⎝ ⎠ ⎟ 

⎜ ⎛ 1⎞ ⎟ ⎜ ⎛4⎞ 

⎟ 

1− − 

⎜ ⎜ ⎟ 3 ⎟ ⎜ ⎜ ⎟ 

3 ⎟ 

⎝ ⎝ ⎠ ⎠ ⎝ ⎝ ⎠ ⎠ 

n 

15 ⎛ ⎛ 1 ⎞ ⎞ 

= 1− − 

4 ⎜ ⎜ ⎟ 

⎝ 3⎠ 

⎟ 

⎝ ⎠ 

23. Neither. There is no common difference or 

common ratio. 

24. 

25. 

26. 

27. 

28. 

3 5 7 9 11 

, , , , , Neither. There is no 

2 4 6 8 10 

common difference or common ratio. 

50 50 

k= 1 k= 

1 

( + ) 

⎛50 50 1 ⎞ 

3k 

= 3 k = 3⎜ 

⎟= 

3825 

⎝ 2 ⎠ 

∑ ∑ 

30 

∑ 

k = 1 

k 

2 

( + )( ⋅ + ) 

30 30 1 2 30 1 

= = 9455 

6 

30 30 30 30 30 

∑ ∑ ∑ ∑ ∑ 

(3k − 9) = 3k− 9 = 3 k− 

9 

k= 1 k= 1 k= 1 k= 1 k= 

1 

⎛30(30 + 1) ⎞ 

= 3⎜ 

⎟−30(9) 

⎝ 2 ⎠ 

= 1395 − 270 = 1125 

40 40 40 

∑ ∑ ∑ 

( − 2k 

+ 8) = − 2k+ 

8 

k= 1 k= 1 k= 

1 

40 40 

∑ 

∑ 

=− 2 k + 8 

k= 1 k= 

1 

⎛40(1 + 40) ⎞ 

=− 2⎜ 

⎟+ 

40(8) 

⎝ 2 ⎠ 

=− 1640 + 320 =−1320 





29. 

7 

∑ 

k = 1 

10 

30. ∑ ( ) 

k = 1 

⎛ 

7 

⎞ ⎛ 

7 

⎞ 

⎛1⎞ ⎛1⎞ 

k ⎜1− 

1 

1 1 

⎜ ⎟ ⎟ ⎜ − 

3 1 

⎜ ⎟ ⎟ 

⎛ ⎞ ⎜ ⎝ ⎠ ⎟ ⎜ ⎝3⎠ 

⎟ 

⎜ ⎟ = = 

⎝3⎠ 3⎜ 1 ⎟ 3⎜ 2 ⎟ 

1− 

⎛ ⎞ 

⎜ 3 ⎟ ⎜ ⎜ ⎟ 

3 ⎟ 

⎝ ⎠ ⎝ ⎝ ⎠ ⎠ 

1⎛ 

1 ⎞ 

= ⎜1− 

⎟ 

2 ⎝ 2187 ⎠ 

1 2186 1093 

= ⋅ = ≈0.49977 

2 2187 2187 

( ) 

⎛ 

10 

k 1− −2 ⎞ 

⎛1−1024 

⎞ 

− 2 = − 2⎜ 

⎟=−2 

⎜ 

⎜ ⎟ 

1 −( −2) ⎟ 3 

⎝ ⎠ 

⎝ ⎠ 

2 

=− ( − 1023) 

= 682 

3 

31. Arithmetic 

a = 3, d = 4, a = a + ( n− 

1) d 

1 n 1 

a 9 = 3 + (9 − 1)4 = 3+ 8(4) = 3+ 32 = 35 


a = 1, d = − 2, a = a + ( n−1) 

d 

1 n 1 

a 8 = 1 + (8−1)( − 2) = 1+ 7( − 2) = 1− 14 = −13 

33. Geometric 

1 

n 1 

a1 = 1, r = , n = 11; an 

= a1r − 

10 

a 

11 

11−1 10 

⎛ 1 ⎞ ⎛ 1 ⎞ 

= 1⋅ ⎜ ⎟ = ⎜ ⎟ 

⎝ 10 ⎠ ⎝ 10 ⎠ 

1 

= 

10,000,000,000 

34. Geometric 

n 

a = 1, r = 2, n= 11; a = a r − 

1 n 1 

11−1 10 

( ) ( ) 

a 11 = 1⋅ 2 = 2 = 1024 


a = 2, d = 2, n = 9, a = a + ( n− 

1) d 

1 n 1 

a 9 = 2 + (9− 1) 2 = 2+ 

8 2 

= 9 2 ≈12.7279 

36. Geometric 

n 

a = 2, r = 2, n= 9, a = a r − 

1 n 1 

− 

( ) ( ) 

9 1 8 

a9 = 2 2 = 2 2 = 2⋅16 

= 16 2 ≈ 22.6274 

1 

1 

37. 7 1 20 1 

a = a + 6d = 31 a = a + 19d 

= 96; 

Solve the system of equations: 

a + 6d 

= 31 

1 

a1 

+ 19d 

= 96 

Subtract the second equation from the first 

equation and solve for d. 

− 13d 

=−65 

d = 5 

a 1 = 31− 6(5) = 31− 30 = 1 

a = a + n− 

d 

n 

( ) 

( n )( ) 

1 1 

= 1+ −1 5 

= 1+ 5n 

−5 

= 5n 

−4 

General formula: { a } = { 5n− 

4} 

38. 8 1 17 1 

n 

a = a + 7d =− 20 a = a + 16d 

= − 47; 


a + 7d 

=−20 

1 

a1 

+ 16d 

=−47 



− 9d 

= 27 

d = −3 

a 1 = −20 −7( − 3) = − 20 + 21 = 1 

a = a + n− 

d 

n 

( ) 

( n )( ) 

1 1 

= 1+ −1 −3 

= 1− 3n 

+ 3 

=− 3n 

+ 4 

General formula: { a } = { − 3n+ 

4} 

39. 10 1 18 1 

a = a + 9d = 0 a = a + 17d 

= 8; 


a + 9d 

= 0 

1 

a1 

+ 17d 

= 8 



− 8d 

=−8 

d = 1 

a 1 = − 9(1) = − 9 

a = a + n− 

d 

n 

( ) 

( n )( ) 

1 1 

=− 9+ −1 1 

=− 9+ n −1 

= n −10 

General formula: { a } = { n− 

10} 

n 

n 





40. 12 1 22 1 

a = a + 11d = 30 a = a + 21d 

= 50 ; 


a1 

+ 11d 

= 30 

a1 

+ 21d 

= 50 



− 10d 

= −20 

d = 2 

a 1 = 30 − 11(2) = 30 − 22 = 8 

an 

= a1 + ( n−1) 

d 

= 8+ ( n −1)( 2) 

= 8+ 2n 

−2 

= 2n 

+ 6 

a = 2n+ 

6 

General formula: { } { } 

1 

41. a1 

= 3, r = 

3 


a1 

3 3 9 

Sn 

= = = = 

1− r ⎛ 1⎞ ⎛2⎞ 

2 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 3 ⎠ ⎝ 3 ⎠ 

1 

42. a1 

= 2, r = 

2 


a1 

2 2 

Sn 

= = = = 4 

1− r ⎛ 1⎞ ⎛1⎞ 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 2 ⎠ ⎝ 2 ⎠ 

1 

43. a1 

= 2, r = − 

2 


S 

n 

a1 2 2 4 

= = = = 

1− r ⎛ ⎛ 1 ⎞⎞ ⎛3 

⎞ 3 

⎜1−⎜− ⎟ ⎜ ⎟ 

2 

⎟ 

⎝ ⎝ ⎠⎠ 

⎝2 

⎠ 

2 

44. a1 

= 6, r = − 

3 


S 

n 

a1 6 6 18 

= = = = 

1− r ⎛ ⎛ 2 ⎞⎞ ⎛5 

⎞ 5 

⎜1−⎜− ⎟ ⎜ ⎟ 

3 

⎟ 

⎝ ⎝ ⎠⎠ 

⎝3 

⎠ 

n 

1 3 

45. a 

1 

= , r = 

2 2 


5 

46. a 1 = 5 , r = − 

4 


1 

47. a1 

= 4, r = 

2 


S 

n 

a1 4 4 

= = = = 8 

1− r ⎛ 1⎞ ⎛1⎞ 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 2 ⎠ ⎝ 2 ⎠ 

3 

48. a1 

= 3, r = − 

4 


S 

49. I: 

50. I: 

n 

a1 3 3 12 

= = = = 

1− r ⎛ ⎛ 3 ⎞⎞ ⎛7 

⎞ 7 

⎜1−⎜− ⎟ ⎜ ⎟ 

4 

⎟ 

⎝ ⎝ ⎠⎠ 

⎝4 

⎠ 

31 ⋅ 

n = 1: 3⋅ 1= 3and (1+ 1) = 3 

2 

3k 

II: If 3+ 6+ 9+ + 3 k = ( k+ 


2 

3+ 6+ 9+ + 3k+ 3( k+ 

1) 

= [ 3+ 6+ 9+ + 3k] 

+ 3( k+ 

1) 

3 k 

= ( k + 1) + 3( k + 1) 

2 

⎛3k 

⎞ 3( k + 1) 

= ( k + 1) ⎜ + 3 ⎟= [( k+ 1) + 1] 

⎝ 2 ⎠ 2 


true. 

2 

n = 1: 4⋅1− 2 = 2 and 2(1) = 2 

II: If 2+ 6+ 10 + + (4k− 2) = 2k 


2 + 6 + 10 + + (4k− 2) + [4( k+ 1) −2] 

= 2 + 6 + 10 + + (4k− 2) + 4k+ 

2 

[ ] 

2 2 

( ) ( ) 2 

= 2k + 4k+ 2= 2 k + 2k+ 1 = 2 k+ 

1 


true. 

2 





51. I: 

11 − 

1 

n = 1: 2⋅ 3 = 2 and 3 − 1= 

2 

k−1 

II: If 2+ 6+ 18+ + 2⋅ 3 = 3 −1, then 

k− 

1 k+ 1−1 

2+ 6+ 18+ + 2⋅ 3 + 2⋅3 

k−1 

k 

= ⎡2 6 18 23 ⎤ 

⎣ 

+ + + + ⋅ 

⎦ 

+ 23 ⋅ 

k k k k+ 

1 

= 3 − 1+ 2⋅ 3 = 3⋅3 − 1= 3 −1 


true. 

11 − 

1 

52. I: n = 1: 3⋅ 2 = 3and 3( 2 − 1) 

= 3 

k−1 

k 

II: If 3+ 6+ 12+ + 3⋅ 2 = 3( 2 −1) 

k 

k− 

1 k+ 1−1 

3+ 6+ 12+ + 32 ⋅ + 32 ⋅ 

k−1 

= ⎡3 6 12 32 ⎤ 

⎣ 

+ + + + ⋅ 

⎦ 

+ 32 ⋅ 

= 32 − 1+ 32 ⋅ 


k 

k 

( ) 

k k k k+ 

1 

( ) ( ) ( ) 

= 3⋅ 2 − 1+ 2 = 3 2⋅2 − 1 = 3 2 −1 


true. 

k 

54. I: 

1 3 2 2 

= ⋅ ⎡6k 6k 9k 9k 2k 

2⎤ 

2 ⎣ 

+ + + + + 

⎦ 

1 2 

= ⋅ ⎡6k ( k+ 1) + 9k( k+ 1) + 2( k+ 

1) 

⎤ 

2 ⎣ 

⎦ 

1 ( 1) 6 2 

k ⎡ k 12 k 6 3 k 3 1 

= ⋅ + + + − − − ⎤ 

2 ⎣ 

⎦ 

1 2 

= ⋅ ( k+ 1) ⎡ 6( k 2 k 1) 3( k 1) 1 ⎤ 

2 ⎣ 

+ + − + − 

⎦ 

1 ( 1) 6( 1) 2 

= ⋅ k+ ⎡ k+ − 3( k+ 1) − 1 ⎤ 

2 ⎣ 

⎦ 


true. 

1 

n = 1: 1(1 + 2) = 3and ⋅ (1+ 1)(2⋅ 1+ 7) = 3 

6 

II: If 

k 

13 ⋅ + 24 ⋅ + + kk ( + 2) = ( k+ 1)(2k+ 

7) , 

6 


13 ⋅ + 24 ⋅ + + kk ( + 2) + ( k+ 1)( k+ 1+ 

2) 

= 13 ⋅ + 24 ⋅ + + kk ( + 2) + ( k+ 1)( k+ 

3) 

[ ] 

53. I: n = 1: 

k 

= ( k + 1)(2 k + 7) + ( k + 1)( k + 3) 

2 1 

6 

2 

(3⋅1− 2) = 1 and ⋅1(6 ⋅1 −3⋅1− 1) = 1 

( k + 1) 

2 

2 

= ( 2 k + 7 k+ 6 k+ 

18 ) 

6 

II: If 

( k + 1) 2 

2 2 2 1 2 

= 

1 + 4 + + (3k− 2) = ⋅k( 6k −3k−1) 

, 

( 2 k + 13 k+ 

18 ) 

6 

2 


( k + 1) 

= [( k + 1) + 1][2( k+ 1) + 7] 

2 2 2 2 

2 

1 + 4 + 7 + + (3k− 2) + ( 3( k+ 1) −2 

6 

) 


⎡ 

2 2 2 2 2 

= 1 + 4 + 7 + + (3k− 2) ⎤ 

⎣ 

 

⎦ 

+ (3k+ 

1) 

true. 

1 2 2 

= ⋅k( 6k −3k− 1 ) + (3k+ 

1) 

⎛5⎞ 5! 54321 ⋅ ⋅ ⋅ ⋅ 54 ⋅ 

2 

55. ⎜ ⎟= = = = 10 

1 

⎝2⎠ 

2!3! 21321 ⋅ ⋅ ⋅ ⋅ 21 ⋅ 

⎡ 

3 2 2 

= ⋅ 6k −3k − k+ 18k + 12k+ 

2⎤ 

2 ⎣ 

⎦ 

⎛8⎞ 1 

8! 87654321 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 87 ⋅ 

⎡ 

3 2 

= ⋅ 6k + 15k + 11k+ 

2⎤ 

56. ⎜ ⎟= = = = 28 

2 ⎣ 

⎦ 

⎝6⎠ 

6!2! 65432121 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 21 ⋅ 

1 ( 1) 6 2 

= ⋅ k+ ⎡ k + 9 k+ 

2 ⎤ 

2 ⎣ ⎦ 

_________________________________________________________________________________________________ 

57. 

⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ 

( x+ 2) = ⎜ ⎟x + ⎜ ⎟x ⋅ 2+ ⎜ ⎟x ⋅ 2 + ⎜ ⎟x ⋅ 2 + ⎜ ⎟x 

⋅ 2 + ⎜ ⎟⋅2 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ 

5 5 4 3 2 2 3 1 4 5 

5 4 3 2 

= x + 5⋅ 2x + 10⋅ 4x + 10⋅ 8x + 5⋅ 16x+ 1⋅32 

5 4 3 2 

= x + 10x + 40x + 80x + 80x+ 

32 





58. 

59. 

60. 

⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ 

( x− 3) = ⎜ ⎟x + ⎜ ⎟x ( − 3) + ⎜ ⎟x ( − 3) + ⎜ ⎟x( − 3) + ⎜ ⎟x 

( −3) 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ 

4 4 3 2 2 3 0 4 

4 3 2 

= x + 4( − 3) x + 6⋅ 9x + 4( − 27) x+ 

81 

4 3 2 

= x − 12x + 54x − 108x+ 

81 

⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ 

(2x+ 3) = ⎜ ⎟(2 x) + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟(2 x) ⋅ 3 + ⎜ ⎟⋅3 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ 

5 5 4 3 2 2 3 1 4 5 

5 4 3 2 

= 32x + 5⋅16x ⋅ 3 + 10⋅8x ⋅ 9 + 10⋅4x ⋅ 27 + 5⋅2x⋅ 81+ 1⋅243 

5 4 3 2 

= 32x + 240x + 720x + 1080x + 810x+ 

243 

⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ ⎛4⎞ 

(3x− 4) = ⎜ ⎟(3 x) + ⎜ ⎟(3 x) ( − 4) + ⎜ ⎟(3 x) ( − 4) + ⎜ ⎟(3 x)( − 4) + ⎜ ⎟( −4) 

⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ 

4 4 3 2 2 3 4 

4 3 2 

= 81x + 4⋅27 x ( − 4) + 6⋅9x ⋅ 16 + 4⋅3 x( − 64) + 1⋅256 

4 3 2 

= 81x − 432x + 864x − 768x+ 

256 

_________________________________________________________________________________________________ 

61. n = 9, j = 2, x = x, a = 2 

⎛9⎞ 

9! 9⋅8 

⎜ ⎟x ⋅ 2 = ⋅ 4x = ⋅ 4x = 144x 

⎝2⎠ 

2!7! 2⋅1 

7 


7 2 7 7 7 

62. n = 8, j = 5, x = x, a =− 3 

⎛8⎞ 

3 5 8! 

3 

⎜ ⎟x 

( − 3) = ( −243) 

x 

⎝5⎠ 

5!3! 

876 ⋅ ⋅ 

= ( − 243) x =− 13,608 x 

321 ⋅ ⋅ 

3 


63. n = 7, j = 5, x = 2 x, a = 1 

3 3 

⎛7⎞ 

7! 7⋅6 

⎜ ⎟(2 x) ⋅ 1 = ⋅ 4 x (1) = ⋅ 4x = 84x 

⎝5⎠ 

5!2! 2⋅1 

2 


2 5 2 2 2 

64. n = 8, j = 2, x = 2 x, a = 1 

⎛8⎞ 

6 2 8! 6 

⎜ ⎟(2 x) ⋅ 1 = ⋅64 x (1) 

⎝2⎠ 

2!6! 

87 ⋅ 

= ⋅ 64 x = 1792 x 

21 ⋅ 

6 


6 6 

65. This is an arithmetic sequence with 

a1 = 80, d =− 3, n= 

25 

a. a 25 = 80 + (25 −1)( − 3) = 80 − 72 = 8 bricks 

25 

S = (80 + 8) = 25(44) = 1100 bricks 

2 

1100 bricks are needed to build the steps. 

b. 25 

66. This is an arithmetic sequence with 

a1 = 30, d =− 1, a n = 15 

15 = 30 + ( n −1)( −1) 

− 15 =− n + 1 

− 16 =−n 

n = 16 

16 

S 16 = (30 + 15) = 8(45) = 360 tiles 

2 

360 tiles are required to make the trapezoid. 


3 

a1 

= 20, r = . 

4 

a. After striking the ground the third time, the 

3 

⎛3⎞ 135 

height is 20⎜ 

⎟ = ≈ 8.44 feet . 

⎝4⎠ 

16 

th 

b. After striking the ground the n time, the 

n 

⎛3 

⎞ 

height is 20 ⎜ ⎟ feet . 

⎝4 

⎠ 





c. If the height is less than 6 inches or 0.5 feet, 

then: 

n 

⎛3 

⎞ 

0.5 ≥ 20⎜ ⎟ 

⎝ 4 ⎠ 

n 

⎛3 

⎞ 

0.025 ≥ ⎜ ⎟ 

⎝4 

⎠ 

⎛3 

⎞ 

log ( 0.025) 

≥ nlog ⎜ ⎟ 

⎝ 4 ⎠ 

log ( 0.025) 

n ≥ 

≈12.82 

⎛3 

⎞ 

log ⎜ ⎟ 

⎝ 4 ⎠ 

The height is less than 6 inches after the 

13th strike. 

d. Since this is a geometric sequence with 

r < 1 , the distance is the sum of the two 

infinite geometric series - the distances 

going down plus the distances going up. 

Distance going down: 

20 20 

S down = = = 80 feet. 

⎛ 3⎞ ⎛1⎞ 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 4 ⎠ ⎝ 4 ⎠ 

Distance going up: 

15 15 

S up = = = 60 feet. 

⎛ 3⎞ ⎛1⎞ 

⎜1− 

⎟ ⎜ ⎟ 

⎝ 4 ⎠ ⎝ 4 ⎠ 

The total distance traveled is 140 feet. 

68. a. Since the interest rate is 6.75% per annum 

compounded monthly, this is equivalent to a 

rate of (6.75/12)% each month. Defining a 


A0 

= 190,000 

⎛ 0.0675 ⎞ 

An 

= ⎜1+ ⎟An 

− 1 −1232.34 


⎛ 0.0675 ⎞ 

⎜1+ ⎟ 190,000 −1232.34 


= $189,836.41 

b. ( ) 




After 252 months, the balance is below 

$100,000. The balance is about $99,540. 


The loan will be paid off after 360 months (or 

30 years). They will make 359 payments of 

$1232.34 plus a last payment of $1221.27 plus 

interest. The total amount paid is: 

⎛ 0.0675 ⎞ 

359(1232.34) + 1221.27⎜1+ 

⎟ 


≈ $443,638.20 



loan: 443,638.20 − 190,000 = $253,638.20 . 

g. (a) Since the interest rate is 6.75% per annum 

compounded monthly, this is equivalent to 

a rate of (6.75/12)% each month. 

Defining a recursive sequence, we have: 

A0 

= 190,000 

⎛ 0.0675 ⎞ 

An 

= ⎜1+ ⎟An 

− 1 −1332.34 


⎛ 0.0675 ⎞ 

⎜1+ ⎟ 190,000 −1332.34 


= $189,736.41 

(b) ( ) 







c. Scroll through the table: 


At the beginning of the 33rd month, or after 

32 payments have been made, the balance is 

below $4,000. The balance is $3982.8. 


At the beginning of the 192nd month, or 

after 191 payments have been made, the 


balance is about $99,288. 


The loan will be paid off after 289 

months. They will make 288 payments 

of $1332.34 plus a last payment of 

$1142.80 plus interest. The total 

amount paid is: 

⎛ 0.0675 ⎞ 

288(1332.34) + 1142.80⎜1+ 

⎟ 


≈ $384,863.15 




384,863.15 − 190,000 = $19,863.15 . 

69. a. b1 = 5,000, bn 

= 1.015bn 

− 1− 

100 

b2 = 1.015b( 2−1) 

− 100 = 1.015b1−100 

= 1.015 5000 − 100 = $4975 

( ) 

b. Enter the recursive formula in Y= and draw 

the graph: 

The balance will be paid off at the end of 94 

months or 7years and 10 months. 

Total payments = 

100(93) + 11.01 1.015 = $9,311.18 

( ) 

e. The total interest expense is the difference 

between the total of the payments and the 

original balance: 

100(93) + 11.18 − 5,000 = $4,311.18 


a1 = 20,000, r = 1.04, n = 5 . Find the fifth term 

of the sequence: 

5−1 4 

a 5 = 20,000(1.04) = 20,000(1.04) = 23,397.17 

Her salary in the fifth year will be $23,397.17. 

Chapter 12 Test 

1. 

2 

n −1 

an 

= 

n + 8 

2 2 

1 −1 0 2 −1 3 

a1 = = = 0, a2 

= = , 

1+ 8 9 2+ 

8 10 

2 2 

3 −1 8 4 −1 15 5 

a3 = = , a4 

= = = , 

3+ 8 11 4+ 

8 12 4 

2 

5 −1 24 

a5 

= = 

5+ 

8 13 

The first five terms of the sequence are 0, 3 

10 , 

8 

11 , 5 24 

, and 

4 13 . 





2. a1 an 

an 

− 1 

= 4; = 3 + 2 

a2 = 3a1+ 2= 3( 4) 

+ 2= 

14 

a3 = 3a2 

+ 2= 3( 14) 

+ 2= 

44 

a4 = 3a3 

+ 2= 3( 44) 

+ 2= 

134 

a = 3a 

+ 2 = 3 134 + 2 = 404 

5 4 

( ) 

The first five terms of the sequence are 4, 14, 44, 

134, and 404. 

3 

3. ∑ ( 1) 

4. 

5. 

k = 1 

k + 1 ⎛k 

+ 1⎞ 

− ⎜ 2 ⎟ 

⎝ k ⎠ 

11 + 1+ 1 21 + 2+ 1 31 + 3+ 

1 

= ( − 1 

⎛ ⎞ 

) ( 1 

⎛ ⎞ 

) ( 1 

⎛ ⎞ 

⎜ ) 

2 

⎟+ − ⎜ 

2 

⎟+ − ⎜ 2 ⎟ 

⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ 

2⎛2⎞ 3⎛3⎞ 4⎛4⎞ 

= ( − 1) ⎜ ( 1) ( 1) 

1 

⎟+ − ⎜ 

4 

⎟+ − ⎜ 

9 

⎟ 

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 

3 4 61 

= 2 − + = 4 9 36 

4 

∑ 

k 

⎡⎛2 

⎞ ⎤ 

⎢⎜ 

− k ⎥ 

3 

⎟ 

⎢⎣⎝ 

⎠ ⎥⎦ 

k = 1 

⎡ 1 ⎤ ⎡ 2 ⎤ ⎡ 3 ⎤ ⎡ 4 

⎤ 

2 2 2 2 

( ) ( ) ( ) ( ) 

= ⎢ − 1 2 3 4 

3 ⎥ + ⎢ − 

3 ⎥ + ⎢ − 

3 ⎥ + ⎢ − 

3 ⎥ 

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 

⎛2 ⎞ ⎛4 ⎞ ⎛ 8 ⎞ ⎛16 

⎞ 

= ⎜ − 1 2 3 4 

3 

⎟+ ⎜ − 

9 

⎟+ ⎜ − 

27 

⎟+ ⎜ − 

81 

⎟ 

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 

1 14 73 308 

=− − − − 

3 9 27 81 

680 

=− 

81 

2 3 4 11 

− + − + ... + 

5 6 7 14 

Notice that the signs of each term alternate, with 

the first term being negative. This implies that 

the general term will include a power of − 1 . 

Also note that the numerator is always 1 more 

than the term number and the denominator is 4 

more than the term number. Thus, each term is in 

k ⎛ k + 1 ⎞ 

the form ( −1) 

⎜ 

k + 4 

⎟. The last numerator is 11 

⎝ ⎠ 

which indicates that there are 10 terms. 

10 

2 3 4 11 k ⎛ k + 1⎞ 

− + − + ... + = ( 1) 

5 6 7 14 

∑ − ⎜ 

k + 4 

⎟ 

⎝ ⎠ 

k = 1 

6. 6,12,36,144,... 

12 − 6 = 6 and 36 − 12 = 24 

The difference between consecutive terms is not 

constant. Therefore, the sequence is not 

arithmetic. 

7. 

12 = 2 and 36 

6 12 

= 3 

The ratio of consecutive terms is not constant. 

Therefore, the sequence is not geometric. 

1 4 

n 

a n = − ⋅ 

2 

1 n 1 n−1 

a − ⋅4 − ⋅4 ⋅4 

n 2 2 

= = = 4 

a 1 n−1 1 n−1 

− ⋅4 − ⋅4 

n−1 2 2 

Since the ratio of consecutive terms is constant, 

the sequence is geometric with common ratio 

1 1 

r = 4 and first term a 1 =− ⋅ 4 =− 2 . 

2 

The sum of the first n terms of the sequence is 

given by 

n 

1− 

r 

Sn 

= a1 

⋅ 

1− 

r 

n 

1− 

4 

=−2⋅ 

1 − 4 

2 

( 1 4 

n 

= − ) 

3 

8. −2, −10, −18, − 26,... 

−10 −( − 2) 

= − 8 , ( ) 

−18 − − 10 = − 8 , 

−26 −( − 18) 

= − 8 


constant. Therefore, the sequence is arithmetic 

with common difference d =− 8 and first term 

a 1 = − 2 . 

an 

= a1 + ( n−1) 

d 

= − 2+ ( n −1)( −8) 

=−2− 8n 

+ 8 

= 6−8n 


given by 

n 

Sn 

= ( a+ 

an) 

2 

n 

= ( − 2+ 6−8n) 

2 

n 

= ( 4−8n) 

2 

= n 2−4n 

( ) 





n 

9. a n =− + 7 

2 

⎡ n ⎤ ⎡ ( n −1) 

⎤ 

an 

− an−1 

= 

⎢ 

− + 7 − 7 

2 

⎢− + 

2 

⎥ 

⎣ 

⎥ 

⎦ ⎣ ⎦ 

n n−1 

=− + 7+ −7 

2 2 

1 

=− 

2 


constant. Therefore, the sequence is arithmetic 

1 

with common difference d =− and first term 

2 

1 13 

a 1 =− + 7 = . 

2 2 


given by 

n 

Sn 

= ( a1 

+ an) 

2 

n⎛13 

⎛ n ⎞⎞ 

= 7 

2⎜ 

+ 

2 

⎜− + 

2 

⎟⎟ 

⎝ ⎝ ⎠⎠ 

n⎛27 

n⎞ 

= 

2 

⎜ − 

2 2 

⎟ 

⎝ ⎠ 

n 

= ( 27 −n) 

4 

10. 

8 

25,10, 4, ,... 

5 

10 2 

= , 4 2 

8 

5 8 1 2 

= , = ⋅ = 

25 5 10 5 4 5 4 5 


Therefore, the sequence is geometric with 

common ratio r = 2 

5 

and first term a 1 = 25 . 


given by 

n 

2 

n 

1− ⎛ ⎞ 

⎜ 

1 

5 

⎟ 

− r 

Sn 

= a1 

⋅ = 25⋅ 

⎝ ⎠ 

1− 

r 2 

1− 

5 

11. 

2n 

− 3 

an 

= 

2n 

+ 1 

2n 

− 3 2( n −1) 

−3 

an 

− an−1 

= − 

2n 

+ 1 2( n − 1) 

+ 1 

2n−3 2n−5 

= − 

2n+ 1 2n−1 

= 

( 2n− 3)( 2n−1) −( 2n− 5)( 2n+ 

1) 

( 2n+ 1)( 2n−1) 

2 2 

( 4n − 8n+ 3) −( 4n −8n−5) 

= 

2 

4n 

−1 

8 

= 

2 

4n 

− 1 

The difference of consecutive terms is not 

constant. Therefore, the sequence is not 

arithmetic. 

2n 

− 3 

an 

= 

2n 

+ 1 

an−1 

2( n −1) 

−3 

2( n − 1) 

+ 1 

2n−3 2n−1 

= ⋅ 

2n+ 1 2n−5 

( 2n− 

3)( 2n−1) 

= 

2n+ 1 2n− 

5 

( )( ) 

The ratio of consecutive terms is not constant. 

Therefore, the sequence is not geometric. 

−64 1 

12. For this geometric series we have r = =− 

256 4 

1 1 

and a 1 = 256 . Since r =− = < 1, the series 

4 4 

converges and we get 

a1 

256 256 1024 

S∞ = 1− 

r 

= 1 5 

1−( − 

4 ) 

= = 5 

4 

n 

⎡ 2 

1 

⎛ ⎞ ⎤ 

⎢ − ⎜ 

5 

⎟ ⎥ 

n 

⎢ ⎝ ⎠ ⎥ 5⎡ 

⎛2⎞ 

⎤ 

= 25⋅ ⎣ ⎦ 

= 25⋅ ⎢1 

− 

3 3 

⎜ 

5 

⎟ ⎥ 

⎢⎣ 

⎝ ⎠ ⎥⎦ 

5 

n 

125 ⎡ ⎛2 

⎞ ⎤ 

= ⎢1 

− 

3 

⎜ ⎥ 

5 

⎟ 

⎢⎣ 

⎝ ⎠ ⎥⎦ 





⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ ⎛5⎞ 

⎜0⎟ ⎜1⎟ ⎜2⎟ ⎜3⎟ ⎜4⎟ ⎜5⎟ 

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 

5 4 3 2 

= 243m + 5⋅81m ⋅ 2 + 10⋅27m ⋅ 4 + 10⋅9m ⋅ 8 + 5⋅3m⋅ 16 + 32 

5 4 3 2 

= 243m + 810m + 1080m + 720m + 240m+ 

32 

5 5 4 3 2 2 3 4 5 

13. ( 3m+ 2) = ( 3m) + ( 3m) ( 2) + ( 3m) ( 2) + ( 3m) ( 2) + ( 3m)( 2) + ( 2) 

14. First we show that the statement holds for n = 1 . 

⎛ 1⎞ 

⎜1+ = 1+ 1= 

2 

1 

⎟ 

⎝ ⎠ 

1 1 1 1 

The equality is true for n = 1 so Condition I holds. Next we assume that ⎜ ⎛ 1+ 1 1 ... 1 n 1 

1 

⎟⎜ ⎞⎛ + ⎞⎛ + ⎞ ⎛ + ⎞ = + 

2 

⎟⎜ 

3 

⎟ ⎜ 

n 

⎟ 

⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠ 

is true for some k, and we show the formula then holds for k + 1. We assume that 

⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞ 

⎜1+ 1 1 ... 1 k 1 

1 

⎟⎜ + + + = + 

2 

⎟⎜ 

3 

⎟ ⎜ 

k 

⎟ . Now we need to show that 

⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠ 

⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎛ 1 ⎞ 

⎜1+ 1 1 ... 1 1 ( k 1) 

1 k 2 

1 

⎟⎜ + 

2 

⎟⎜ + + + = + + = + 

3 

⎟ ⎜ 

k 

⎟⎜ 

k+ 

1 

⎟ 

. 

⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ 

We do this as follows: 

⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎛ 1 ⎞ ⎡⎛ 1⎞⎛ 1⎞⎛ 1⎞ ⎛ 1⎞⎤⎛ 1 ⎞ 

⎜1+ 1 1 ... 1 1 1 1 1 ... 1 1 

1 

⎟⎜ + 

2 

⎟⎜ + 

3 

⎟ ⎜ + 

k 

⎟⎜ + 

k 1 

⎟ = ⎜ + 

1 

⎟⎜ + 

2 

⎟⎜ + 

3 

⎟ ⎜ + 

k 

⎟ ⎜ + 

k 1 

⎟ 

⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ + ⎠ 

⎢ 

⎥ 

⎣⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎦⎝ + ⎠ 

⎛ 1 ⎞ 

= ( k + 1) 

⎜1 + (using the induction assumption) 

k 1 

⎟ 

⎝ + ⎠ 

1 

= ( k+ 1) ⋅ 1+ ( k+ 1) 

⋅ 

k + 1 

= k + 1+ 

1 

= k + 2 

Condition II also holds. Thus, formula holds true for all natural numbers. 

_________________________________________________________________________________________________ 

15. The yearly values of the Durango form a 

geometric sequence with first term a 1 = 31,000 

and common ratio r = 0.85 (which represents a 

15% loss in value). 

1 

a 31,000 ( 0.85) 

n − 

n = ⋅ 

The nth term of the sequence represents the 

value of the Durango at the beginning of the nth 

year. Since we want to know the value after 10 

years, we are looking for the 11 th term of the 

sequence. That is, the value of the Durango at the 

beginning of the 11 th year. 

11 1 

a ( ) 10 

11 = a1 ⋅ r − = 31,000 ⋅ 0.85 = 6,103.11 

After 10 years, the Durango will be worth 

$6,103.11. 

16. The weights for each set form an arithmetic 

sequence with first term a 1 = 100 and common 

difference d = 30 . If we imagine the weightlifter 

only performed one repetition per set, the total 

weight lifted in 5 sets would be the sum of the 

first five terms of the sequence. 

an 

= a1 

+ ( n−1) 

d 

a5 

= 100 + ( 5 − 1)( 30) = 100 + 4( 30) 

= 220 

n 

Sn 

= ( a+ 

an) 

2 

S 5 

5 = ( 100 + 220) = 5 

2 2( 320) 

= 800 

Since he performs 10 repetitions in each set, we 

multiply the sum by 10 to obtain the total weight 

lifted. 

10( 800) 

= 8000 

The weightlifter will have lifted a total of 8000 

pounds after 5 sets. 





Chapter 12 Cumulative Review 

1. 

2. a. 

2 

x = 9 

2 2 

x = 9 or x =−9 

x =± 3 or x =± 3i 

The solution set is { −3,3, − 3,3} i i . 

b. 

x 

2 2 

+ y = 100 and 

2 

y = 3x 

. 

2 2 

⎪ 

⎧ x + y = 

3 

⎯⎯→ 

2 

x + 

2 

y = 

2 2 

100 3 3 300 

⎨ 

⎪⎩ 

y = 3 x ⎯⎯→− 3 x + y = 0 

2 

3y 

+ y = 300 

2 

2 

3y 

+ y− 300= 

0 

( )( ) 

( ) 

− 1± 1 −4 3 −300 − 1± 

3601 

y = = 

23 6 

Substitute and solve for x: 

− 1+ 3601 2 − 1+ 

3601 

y = ⇒ 3x 

= 

6 6 

2 − 1+ 3601 − 1+ 

3601 

x = ⇒ x =± 

18 18 

or 

−1− 

3601 2 −1− 

3601 

y = ⇒ 3x 

= 

6 6 

2 −1− 

3601 −1− 

3601 

x = ⇒ x = ± 

18 18 

−1− 

3601 

which is not real since < 0 

18 

Therefore, the system has solutions: 

⎧⎛ 

⎪ − 1+ 3601 − 1+ 

3601 

⎞ 

⎨ ⎜ 

, ⎟ , 

⎪ ⎜ 18 6 ⎟ 

⎩ ⎝ 

⎠ 

⎛ 

⎜− 

⎜ 

⎝ 

− 1+ 3601 − 1+ 

3601 ⎞⎫ 

, ⎟ ⎪ ⎬ 

18 6 ⎟ 

⎠⎭ ⎪ 

c. The graphs of the circle and parabola 

intersect at the points 

⎛ − 1+ 3601 − 1+ 

3601 ⎞ 

⎜ 

, ⎟≈ 1.81,9.84 

⎜ 18 6 ⎟ 

⎝ 

⎠ 

⎛ 

− 

⎜ 

⎝ 

x 

3. 2e = 5 

x 5 

e = 

2 

x ⎛5 

⎞ 

ln ( e ) = ln ⎜ ⎟ 

⎝ 2 ⎠ 

⎛5 

⎞ 

x = ln ⎜ ⎟≈0.916 

⎝2 

⎠ 

− 1+ 3601 − 1+ 

3601 ⎞ 

, ⎟≈ − 

18 6 ⎟ 

⎠ 

⎧ ⎛5 

⎞⎫ 

The solution set is ⎨ln ⎜ ⎟⎬ 

⎩ ⎝ 2 ⎠⎭ . 

( ) 

( 1.81,9.84) 

4. slope = m = 5 ; Since the x-intercept is 2, we 

know the point ( 2,0 ) is on the graph of the line 

and is a solution to the equation y = 5x+ b. 

y = 5x+ 

b 

0= 5( 2) 

+ b 

0= 10+ 

b 

− 10 = b 

Therefore, the equation of the line with slope 5 

and x-intercept 2 is y = 5x− 10. 

5. Given a circle with center (–1, 2) and containing 

the point (3, 5), we first use the distance formula 

to determine the radius. 

( ( )) ( ) 

2 2 2 2 

r = 3− − 1 + 5− 2 = 4 + 3 

= 16 + 9 = 25 = 5 

Therefore, the equation of the circle is given by 

2 2 2 

x− − 1 + y− 2 = 5 

( ( )) ( ) 

( x ) ( y ) 

2 2 2 

+ 1 + − 2 = 5 

2 2 

x + 2x+ 1+ y − 4y+ 4= 

25 

2 2 

x + y + 2x−4y− 20= 

0 




Chapter 12 Cumulative Review 

3x 

x 2 

6. f ( x) 

= , g ( x) = 2x+ 

1 

− 

a. g ( ) ( ) 

f ( ) 

( ) 

2 = 2 2 + 1= 

5 

35 15 

5 = = = 5 

5− 

2 3 

( f g)( ) f ( g( )) f ( ) 

b. f ( ) 

2 = 2 = 5 = 5 

( ) 

34 12 

4 = = = 6 

4− 

2 2 

g ( 6) = 2( 6) 

+ 1= 

13 

( g 

f )( ) g( f ( )) g( ) 

4 = 4 = 6 = 13 

c. ( f g)( x) = f ( g( x) 

) 

32 ( x + 1) 

= 

( x + ) − 

2 1 2 

6x 

+ 3 

= 

2x 

−1 

d. To determine the domain of the composition 

( f g)( x) 

, we start with the domain of g 

and exclude any values in the domain of g 

that make the composition undefined. 

g x is defined for all real numbers and 

( ) 

( f g)( x) 

is defined for all real numbers 

except 

1 

x = . Therefore, the domain of the 

2 

⎧ 1 ⎫ 

composite ( f g)( x) 

is 

e. ( g 

f )( x) 

⎛ 3x 

⎞ 

= 2⎜ 

⎟+ 

1 

⎝ x − 2 ⎠ 

6x 

= + 1 

x − 2 

6x+ x−2 

= 

x − 2 

7x 

− 2 

= 

x − 2 

⎨xx≠ 

⎬ 

⎩ 2⎭ . 

f. To determine the domain of the composition 

( g f )( x) 

, we start with the domain of f 

and exclude any values in the domain of f 

that make the composition undefined. 

f ( x ) is defined for all real numbers except 

2 g f x is defined for all real 

x = and ( )( ) 

numbers except x = 2 . Therefore, the 

domain of the composite ( g f )( x) 

is 

{ x| x ≠ 2} 

. 

g. g ( x) = 2x+ 

1 

1 

2 

y = 2x+ 

1 

x = 2y 

+ 1 

x − 1= 

2y 

( x 1) 

− = y 

− 

g ( x) = ( x− 

1) 

2 

−1 

The domain of g ( x) 

1 1 

numbers. 

3x 

f x = 

x − 2 

3x 

y = 

x − 2 

3y 

x = 

y − 2 

x( y− 2) 

= 3y 

xy− 2x = 3y 

xy− 3y = 2x 

y( x− 3) 

= 2x 

2x 

y = 

x − 3 

−1 2x 

f ( x) 

= 

x − 3 

f 

h. ( ) 

−1 

The domain of ( x) 

is the set of all real 

is { x| x ≠ 3} 

. 

7. Center: (0, 0); Focus: (0, 3); Vertex: (0, 4); 

Major axis is the y-axis; a = 4; c = 3 . 

2 2 2 

Find b: b = a − c = 16 − 9 = 7 ⇒ b= 

7 

Write the equation using rectangular coordinates: 

2 2 

x y 

+ = 1 

7 16 

Parametric equations for the ellipse are: 

x = 7cos π t ; y = 4sin π t ; 0≤t 

≤ 2 

( ) ( ) 





8. The focus is ( − 1, 3) 

and the vertex is ( 1, 2) 

− . 

Both lie on the vertical line x =− 1. We have 

a = 1 since the distance from the vertex to the 

−1, 3 is above 

focus is 1 unit, and since ( ) 

( − 1, 2) 

, the parabola opens up. The equation of 

the parabola is: 

2 

( x− h) = 4a( y−k) 

2 

( x−( − 1) 

) = 4⋅1⋅( y−2) 

2 

( x+ 1) = 4( y−2) 

9. Center point (0, 4); passing through the pole 

(0,4) implies that the radius = 4 using rectangular 

coordinates: 

2 2 2 

( x − h) + ( y− k) 

= r 

2 2 2 

x− 0 + y− 4 = 4 

10. 

( ) ( ) 

2 

x ( y ) 

+ − 4 = 16 

2 2 

x + y − 8y+ 16= 

16 

2 2 

x + y − 8y 

= 0 

converting to polar coordinates: 

2 

r − 8rsinθ 

= 0 

2 

r = 8rsinθ 

r = 8sinθ 

2 

2 

2sin x−sin x− 3= 0, 0≤ x≤ 

2π 

( x )( x ) 

2sin − 3 sin + 1 = 0 

3 

2sin x− 3 = 0 ⇒ sin x = , which is impossible 

2 

3π 

sin x+ 1 = 0 ⇒ sin x = −1⇒ x = 

2 

⎧3π 

⎫ 

The solution set ⎨ ⎬ 

⎩ 2 ⎭ . 

−1 

11. cos ( − 0.5) 

We are finding the angle θ, −π ≤θ ≤ π, 

whose 

cosine equals − 0.5 . 

cosθ =−0.5 

−π ≤θ ≤π 

2π 

− 1 2π 

θ = ⇒cos ( − 0.5) 

= 

3 3 


1 

sin θ = , θ is in Quadrant II 

4 

a. θ is in Quadrant II ⇒ cosθ 

< 0 

b. 

⎞ 

cosθ 

=− 1− sin θ =− 1− ⎜ ⎟ 

⎝4 

⎠ 

1 15 15 

=− 1− =− =− 

16 16 4 

2 ⎛1 

⎛1 

⎞ 

sinθ 

⎜ ⎟ 

4 ⎛1 ⎞⎛ 4 ⎞ 

tanθ 

= = 

⎝ ⎠ 

= ⎜ ⎟⎜ − ⎟ 

cosθ 

⎛ 15 ⎞ ⎝4 ⎠⎝ 15 ⎠ 

⎜ 

− 

4 ⎟ 

⎝ ⎠ 

1 15 15 

=− ⋅ =− 

15 15 15 

c. sin(2 θ ) = 2sinθcosθ 

⎛1⎞⎛ 15⎞ 

= 2⎜ ⎜− 

⎝ 4 

⎟⎜ 4 ⎟ 

⎠⎝ ⎠ 

15 

=− 

8 

d. 

e. 

2 2 

cos(2 θ) = cos θ −sin 

θ 

2 2 

⎛ 15 ⎞ ⎛1 

⎞ 

= ⎜ 

− 4 ⎟ 

−⎜ ⎟ 

⎝ ⎠ ⎝4⎠ 

15 1 14 7 

= − = = 

16 16 16 8 

π π θ π 

< θ < π ⇒ < < 

2 4 2 2 

θ 

⎛θ 

⎞ 

⇒ is in Quadrant I ⇒ sin ⎜ ⎟> 

0 

2 

⎝2 

⎠ 

⎛ 15 ⎞ 

1− θ 1 cosθ 

⎜ 

− 

4 ⎟ 

⎛ ⎞ − 

sin 

⎝ ⎠ 

⎜ ⎟= = 

⎝ 2 ⎠ 2 2 

⎛4+ 

15⎞ 

⎜ 

4 ⎟ 

4+ 

15 

= 

⎝ ⎠ 

= 

2 8 

4+ 

15 

= 

2 2 

2 




Chapter 12 Projects 

Chapter 12 Projects 

Project I 

Answers will vary based on the year that is used. Data 

used in these solutions will be from 2001. 

1. Race Birth rate Death rate 

White 0.0118 0.00991 

African-American 0.116 0.007758 

American Indian/ 

Alaska native 

0.0137 0.00392 

Asian/Pacific 

Islander 

0.0164 0.003047 

Hispanic 0.0230 0.00306 

Total 0.141 0.008485 

I = net immigration = 973,206 

Population for 2001 = 285,545,000 

2. r = 0.0141 – 0.008485 = 0.005615 

3. pn 

= (1 + 0.005615) pn−1 

+ 973206 

pn 

= (1.005615) pn−1 

+ 973206 

p = 285545000 

0 

4. p1 p0 

= (1.005615) + 973206 

p1 

= (1.005615)(285545000) + 973206 

p = 288,121,541 

1 

The population is predicted to be 288,121,541 in 

2002. 

5. Actual population in 2002: 288,600,000. The 

formula’s prediction was lower but fairly close. 

6. Birth rate: 47.52 per 1000 population (0.04752) 

Death rate: 17.97 per 1000 population (0.01797) 

Population for 2001: 23,985,712 

I = net immigration = − 6956 

r = 0.04752 − 0.01797 = 0.02955 

pn 

= (1 + 0.02955) pn−1 

−6956 

pn 

= (1.02955) pn−1 

−6956 

p0 

= 23,985,712 

p1 = (1.02955) p0 

−6956 

p1 

= (1.02955)(23985712) −6956 

p1 

= 24,687,534 

The population is predicted to be 24,687,534 in 

2002. 

Actual population in 2002: 24,699,073. 

The formula’s prediction was higher but fairly 

close. 

7. Answers will vary. This appears to support the 

article. The growth rate for the U.S. is much 

smaller than the growth rate for Uganda. 

8. It could be but one must consider trends in each 

of the pieces of data to find if the growth rate is 

increasing or decreasing over time. The same 

thing must be examined with respect to the net 

immigration. 

Project II 

1. 2, 4, 8, 16, 32, 64 

2. length n 2 n levels 

This is a geometric sequence: a 2 n 

n = 

Recursive expression: an 

= 2 an−1, a0 

= 1 

3. 

n 

256 = 2 

8 n 

2 = 2 

n = 8 

Project III 

1. Qst = − 3+ 2 Pt− 

1, Qdt = 18−3Pt 

P0 

= 2, b= 2, d = 3, c = 18 

− a = −3 → a = 3 

2. 

3+ 18−2P 

21−2P 

Pt 

= = 

3 3 

2 

Pt 

= 7 − Pt− 

1, P0 

= 2 

3 

−10 

t−1 t−1 

7 

−9 

P t 

10 

Pt 

− 1 





3. P1 

= 

Q 

Q 

17 

3 

=− 3+ 2(2) Q 

⎛17 

⎞ 

= 18−3 ⎜ ⎟ 

⎝ 3 ⎠ 

= 1 Q = 1 

s1 d1 

s1 d1 

29 

P2 

= 

9 

⎛17 ⎞ ⎛29 

⎞ 

Qs2 =− 3+ 2⎜ ⎟ Qd2 

= 18−3⎜ ⎟ 

⎝ 3 ⎠ ⎝ 9 ⎠ 

25 25 

Qs2 = Qd2 

= 

3 3 

The market (supply and demand) are getting 

closer to being the same. 

4. The equilibrium price is 4.20. 

5. It takes 17 time periods. 

6. Qd17 

Q 

s17 

Project IV 

= 18 − 3(4.20) = 5.40 

=− 3+ 2(4.20) = 5.40 

The equilibrium quantity is 5.4. 

1. 1, 2, 4, 7, 11, 16, 22, 29 

2. It is not arithmetic because there is no common 

difference. It is not geometric because there is no 

common ration. 

3. 15 

−2 

−2 

4. y = 2.5x− 

2.5 

The graph does not pass through any of the 

points. 

y6 

= 12.5 

y7 

= 15 

y = 17.5 

8 

6 

5. 

y1 

= 0 

y2 

= 2.5 

y3 

= 5 

y4 

= 7.5 

y = 10 

5 

5 

∑ 

i= 

1 

( y − y ) 

r 

i 

i 

= (0 − 1) + (2.5 − 2) + (5 − 4) + (7.5 − 7) + (10 −11) 

=−1 

This is the sum of the errors. 

2 

y = 0.5x − 0.5x+ 

1 

The graph passes through all of the points. 

y6 

= 16 

y7 

= 22 

y = 29 

8 

y1 

= 1 

y2 

= 2 

y3 

= 4 

y4 

= 7 

y = 11 

5 

5 

∑ 

i= 

1 

( y − y ) = 0 

r 

i 

i 

The sum of the errors is zero. 

6. When trying to obtain the cubic and quartic 

polynomials of best fit, the cubic and quartic 

terms have coefficient zero and the polynomial 

of best fit is given as the quadratic in part e. 

For the exponential function of best fit, 

(0.59)(1.83) x 

y = . 

y6 = 22.2 y7 = 40.6 y8 

= 74.2 

The sum of these errors becomes quite large. 

This error shows that the function does not fit the 

data very well as x gets larger. 

7. The quadratic function is best. 

8. The data does not appear to be either logarithmic 

or sinusoidal in shape, so it does not make sense 

to try to fit one of those functions to the data.

Chapter 12 Sequences; Induction; the Binomial Theorem

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