Chapter 12 Sequences; Induction; the Binomial Theorem

Chapter 12: Sequences; Induction; the Binomial Theorem
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
61. Answers may vary. One possibility follows:
20
1+ 2+ 3+ + 20= ∑ k
k = 1
62. Answers may vary. One possibility follows:
8
3 3 3 3 3
1 + 2 + 3 + + 8 = ∑ k
k = 1
63. Answers may vary. One possibility follows:
1 2 3 13
13
k
+ + + + = ∑
2 3 4 13+ 1 k + 1
k = 1
64. Answers may vary. One possibility follows:
12
1+ 3 + 5 + 7 + + 2(12) − 1 = (2k
−1)
[ ]
n
∑
k = 1
65. Answers may vary. One possibility follows:
1 1 1
6
6 ⎛ 1 ⎞ k ⎛ 1 ⎞
1 − + − + + ( − 1) ⎜ ( 1)
3 9 27
6 ⎟= ∑ − ⎜ k ⎟
⎝3 ⎠ ⎝3
⎠
k = 0
66. Answers may vary. One possibility follows:
11
2 4 8
11
11+ 1 ⎛2⎞ k + 1 ⎛2⎞
− + + + ( − 1) ⎜ ⎟ = ∑ ( −1)
⎜ ⎟
3 9 27
⎝3⎠ ⎝3⎠
k = 1
k
67. Answers may vary. One possibility follows:
2 3
n k
3 3 3 3
3 + n
2 + 3
+ + n
= ∑ k
k = 1
68. Answers may vary. One possibility follows:
1 2 3 n
n
k
+ + + + =
e
2 3
n
∑
k
e e e e
k = 1
69. Answers may vary. One possibility follows:
n
a+ ( a+ d) + ( a+ 2 d) + + ( a+ nd) = ( a+
kd)
n
∑
k = 1
( )
or = ( a+ k −1 d)
∑
k = 0
70. Answers may vary. One possibility follows:
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
( )
1238
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