Chapter 12 Sequences; Induction; the Binomial Theorem

Chapter 12 Review Exercises
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
Conditions I and II are satisfied; the statement is
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 ⋅
, then
k
k
( )
k k k k+
1
( ) ( ) ( )
= 3⋅ 2 − 1+ 2 = 3 2⋅2 − 1 = 3 2 −1
Conditions I and II are satisfied; the statement is
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 ⎣
⎦
Conditions I and II are satisfied; the statement is
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
then
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
then
( k + 1)
= [( k + 1) + 1][2( k+ 1) + 7]
2 2 2 2
2
1 + 4 + 7 + + (3k− 2) + ( 3( k+ 1) −2
6
)
Conditions I and II are satisfied; the statement is
⎡
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
1277
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.