Chapter 12 Sequences; Induction; the Binomial Theorem
Chapter 12 Sequences; Induction; the Binomial Theorem
Chapter 12 Sequences; Induction; the Binomial Theorem
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Section <strong>12</strong>.4: Ma<strong>the</strong>matical <strong>Induction</strong><br />
5. I:<br />
6. I:<br />
1<br />
n = 1: 3⋅1− 1= 2 and ⋅1(3⋅ 1+ 1) = 2<br />
2<br />
1<br />
II: If 2+ 5+ 8 + + (3k− 1) = ⋅ k(3k+<br />
1) , <strong>the</strong>n<br />
2<br />
2 + 5 + 8 + + (3k− 1) + [3( k+ 1) −1]<br />
= [ 2+ 5+ 8 + + (3k− 1) ] + (3k+<br />
2)<br />
1 3 2 1<br />
= ⋅ k(3k+ 1) + (3k+ 2) = k + k+ 3k+<br />
2<br />
2 2 2<br />
3 2 7 1 2<br />
= k + k+ 2= ( 3k + 7k+<br />
4)<br />
2 2 2<br />
= 1 ( k+ 1)(3 k+<br />
4)<br />
2<br />
1<br />
= ( k+ 1)[3( k + 1) + 1]<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
1<br />
n = 1: 3⋅1− 2 = 1and ⋅1(3⋅1− 1) = 1<br />
2<br />
1<br />
II: If 1+ 4+ 7 + + (3k− 2) = ⋅k(3k−1)<br />
,<br />
2<br />
<strong>the</strong>n<br />
1+ 4+ 7 + + (3k− 2) + [3( k+ 1) −2]<br />
= [ 1+ 4+ 7 + + (3k− 2) ] + (3k+<br />
1)<br />
1 3 2 1<br />
= ⋅k(3k − 1) + (3k+ 1) = k − k+ 3k+<br />
1<br />
2 2 2<br />
3 2 5 1 2<br />
= k + k + 1= ( 3k + 5k+<br />
2)<br />
2 2 2<br />
= 1 ( k+ 1)(3 k+<br />
2)<br />
2<br />
1<br />
= ( k+ 1)[3( k + 1) − 1]<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
7. I:<br />
11 −<br />
1<br />
n = 1: 2 = 1and 2 − 1=<br />
1<br />
2 k−1<br />
II: If 1+ 2+ 2 + + 2 = 2 −1, <strong>the</strong>n<br />
2 k−<br />
1 k+ 1−1<br />
1+ 2+ 2 + + 2 + 2<br />
2 k−1<br />
= ⎡1 2 2 2 ⎤<br />
⎣<br />
+ + + +<br />
⎦<br />
+ 2<br />
k k k<br />
= 2 − 1+ 2 = 2⋅2 −1<br />
k + 1<br />
= 2 −1<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
k<br />
k<br />
8. I:<br />
11 − 1 1<br />
n = 1: 3 = 1and (3 − 1) = 1<br />
2<br />
2 k−1 1<br />
II: If 1+ 3+ 3 + + 3 = ⋅(3 −1)<br />
, <strong>the</strong>n<br />
2<br />
2 k−<br />
1 k+ 1−1<br />
1+ 3+ 3 + + 3 + 3<br />
2 k−1<br />
k<br />
= ⎡1 3 3 3 ⎤<br />
⎣<br />
+ + + +<br />
⎦<br />
+ 3<br />
1 k k 1 k 1 k<br />
= ⋅(3 − 1) + 3 = ⋅3 − + 3<br />
2 2 2<br />
3 k 1 1 k<br />
= ⋅3 − = ⋅( 3⋅3 −1)<br />
2 2 2<br />
1 1<br />
( 3<br />
k +<br />
= − 1 )<br />
2<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
11 −<br />
1<br />
9. I: n<br />
1<br />
( )<br />
= 1: 4 = 1and ⋅ 4 − 1 = 1<br />
3<br />
2 k−1 1 k<br />
II: If 1+ 4+ 4 + + 4 = ⋅( 4 −1)<br />
2 k−1<br />
( )<br />
( )<br />
3<br />
2 k−<br />
1 k+ 1−1<br />
1+ 4+ 4 + + 4 + 4<br />
= 1+ 4+ 4 + + 4 + 4<br />
1 1 1<br />
= ⋅ 4 − 1 + 4 = ⋅4 − + 4<br />
3 3 3<br />
4 k 1 1 k<br />
= ⋅4 − = ( 4⋅4 −1)<br />
3 3 3<br />
1 1<br />
( 4<br />
k +<br />
= ⋅ − 1 )<br />
3<br />
k k k k<br />
k<br />
k<br />
, <strong>the</strong>n<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
11 −<br />
1<br />
10. I: n<br />
1<br />
( )<br />
= 1: 5 = 1and ⋅ 5 − 1 = 1<br />
4<br />
2 k−1 1 k<br />
II: If 1+ 5+ 5 + + 5 = ⋅( 5 −1)<br />
4<br />
2 k−<br />
1 k+ 1−1<br />
1+ 5+ 5 + + 5 + 5<br />
2 k−1<br />
k<br />
= ⎡1 5 5 5 ⎤<br />
⎣<br />
+ + + +<br />
⎦<br />
+ 5<br />
1 1 1<br />
= ⋅( 5 − 1)<br />
+ 5 = ⋅5 − + 5<br />
4 4 4<br />
5 k 1 1 k<br />
= ⋅5 − = ( 5⋅5 −1)<br />
4 4 4<br />
1 1<br />
( 5<br />
k +<br />
= ⋅ − 1 )<br />
4<br />
k k k k<br />
, <strong>the</strong>n<br />
Conditions I and II are satisfied; <strong>the</strong> statement is<br />
true.<br />
<strong>12</strong>63<br />
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