+ n/b 8 78. k UI k1 53. 0.00952 0.00952952. . In ((k +1)k) (In k) In (k + 1) Telescoping series For the following telescoping series, find a atmate lim S to obtain the value of the series or state that the series 0 125=1.252 6L k 2 5.1283=5.12838383. . . pamula for the nth term of the sequence of partial sums {S. Then 59-1 3 81. I k +1 k+2 3 I 3 I 00 S0dae I 83. 5 : +2 I 20 (k +6) (k + 7) 85. E) 25k2+ 15k- 4 I * 1=} 87. Explain why or w k=3 (4k - 3)(4k 1) ments are true and 4k2 32k 63 3 E--Y 00 k + 1 61. In k . Ifa conver InJI k 1 I 00 k=1 (3k+ 1)(3k+ 4) 00 b. If a diverg k 1 0=1 00 c. If a conve d. If p diver (2k - 1)(2k + 1 where p is a positive integer I real number -k 00 (k+ p)(k + p 1) 00 I=X e. I where a is a positive integer 00 ә К» ak +1)(ak + a + 1)' f. If the series [=\ I Vk+3 bconver 00 T Vk1 k 1 g. Viewed as a 00 67. 00 9 +2k k25k + 4 9 takes on all =1 69. (tan(k 1) - tan k) (- w 00 88-89. Binary nun 00 k-1 base-10 or decimal k 1 numbers internally Evaluating an infinite series two ways Evaluate the series 0's and l's. For thi I 2k 2k+1 00 I+yC two ways form 0.b,b,b3 The base-10 repre a. Use a telescoping series argument. b. Use a geometric series argument with Theorem 10.8. 71. Evaluating infinite series an infinite series two ways Evaluate the series 88. Verify that th + Ms M TAU 456456456. sin 125=1.252525. 5.12838383. 4 Telescoping series For the following telescoping series, find a e lim S to obtain the value of the series or state that the series 53. 0.00952 0.00952952. In ((k 1)k 6 (In k) In (k + E8275 k 2 for the nth term of the sequence of partial sums S. Then o0 81. 20 hat I k + 1 I k +2 C0 00 I 83. 9 I ¥ k + 3 +2 57. I 0 85. (L(9) .) 25k215k 4 00 10 87. Explain why or (4k 3) (4k 1) ments are true a = 1 4k2+32k + 63 30* E-F! k + 1 61. In 00 00 a. If a con I k 1 k-1 ( 3k + 1)(3k + 4) 00 b. If a, div k 1 0=} 2 00 (2k-1)(2k 1) c. If a co E 7 where p is a positive integer d. If p di I CC real numb . (I + d+ y)(d + 00 [=} e. I where a is a positive integer #2 ak + 1)(ak + a + 1)' 1=\ f. If the seri 1 I Vk+ 1 00 C0 Vk+ 3 k 1 ( g. Viewed: O0 9 k2+5k + 4 99 K+2k takes on 3 = \ 00 69. (tan(k + 1) - tan k) 88-89. Binary k 1 base-10 or deci Evaluating an infinite series two ways Evaluate the series numbers interne 0's and l's. For I two ways. form 0.b,b,b3 The base-10 re 2 2k+1 . => 2 Use a telescoping series argument. b. Use a geometric series argument with Theorem 10.8. infinite series

+ n/b 8 78. k UI k1 53. 0.00952 0.00952952. . In ((k +1)k) (In k) In (k + 1) Telescoping series For the following telescoping series, find a atmate lim S to obtain the value of the series or state that the series 0 125=1.252 6L k 2 5.1283=5.12838383. . . pamula for the nth term of the sequence of partial sums {S. Then 59-1 3 81. I k +1 k+2 3 I 3 I 00 S0dae I 83. 5 : +2 I 20 (k +6) (k + 7) 85. E) 25k2+ 15k- 4 I * 1=} 87. Explain why or w k=3 (4k - 3)(4k 1) ments are true and 4k2 32k 63 3 E--Y 00 k + 1 61. In k . Ifa conver InJI k 1 I 00 k=1 (3k+ 1)(3k+ 4) 00 b. If a diverg k 1 0=1 00 c. If a conve d. If p diver (2k - 1)(2k + 1 where p is a positive integer I real number -k 00 (k+ p)(k + p 1) 00 I=X e. I where a is a positive integer 00 ә К» ak +1)(ak + a + 1)' f. If the series [=\ I Vk+3 bconver 00 T Vk1 k 1 g. Viewed as a 00 67. 00 9 +2k k25k + 4 9 takes on all =1 69. (tan(k 1) - tan k) (- w 00 88-89. Binary nun 00 k-1 base-10 or decimal k 1 numbers internally Evaluating an infinite series two ways Evaluate the series 0's and l's. For thi I 2k 2k+1 00 I+yC two ways form 0.b,b,b3 The base-10 repre a. Use a telescoping series argument. b. Use a geometric series argument with Theorem 10.8. 71. Evaluating infinite series an infinite series two ways Evaluate the series 88. Verify that th + Ms M TAU 456456456. sin 125=1.252525. 5.12838383. 4 Telescoping series For the following telescoping series, find a e lim S to obtain the value of the series or state that the series 53. 0.00952 0.00952952. In ((k 1)k 6 (In k) In (k + E8275 k 2 for the nth term of the sequence of partial sums S. Then o0 81. 20 hat I k + 1 I k +2 C0 00 I 83. 9 I ¥ k + 3 +2 57. I 0 85. (L(9) .) 25k215k 4 00 10 87. Explain why or (4k 3) (4k 1) ments are true a = 1 4k2+32k + 63 30* E-F! k + 1 61. In 00 00 a. If a con I k 1 k-1 ( 3k + 1)(3k + 4) 00 b. If a, div k 1 0=} 2 00 (2k-1)(2k 1) c. If a co E 7 where p is a positive integer d. If p di I CC real numb . (I + d+ y)(d + 00 [=} e. I where a is a positive integer #2 ak + 1)(ak + a + 1)' 1=\ f. If the seri 1 I Vk+ 1 00 C0 Vk+ 3 k 1 ( g. Viewed: O0 9 k2+5k + 4 99 K+2k takes on 3 = \ 00 69. (tan(k + 1) - tan k) 88-89. Binary k 1 base-10 or deci Evaluating an infinite series two ways Evaluate the series numbers interne 0's and l's. For I two ways. form 0.b,b,b3 The base-10 re 2 2k+1 . => 2 Use a telescoping series argument. b. Use a geometric series argument with Theorem 10.8. infinite series

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Can you help me on #61

+
n/b
8
78.
k
UI
k1
53. 0.00952
0.00952952. .
In ((k +1)k)
(In k) In (k + 1)
Telescoping series For the following telescoping series, find a
atmate lim S to obtain the value of the series or state that the series
0
125=1.252
6L
k 2
5.1283=5.12838383. . .
pamula for the nth term of the sequence of partial sums {S. Then
59-1
3
81.
I
k +1 k+2
3
I
3
I
00
S0dae
I
83.
5
:
+2
I
20
(k +6) (k + 7)
85.
E)
25k2+ 15k- 4
I *
1=}
87. Explain why or w
k=3 (4k - 3)(4k 1)
ments are true and
4k2 32k 63
3
E--Y
00
k + 1
61. In
k
. Ifa conver
InJI
k 1
I
00
k=1
(3k+ 1)(3k+ 4)
00
b. If a diverg
k 1
0=1
00
c. If a conve
d. If p diver
(2k - 1)(2k + 1
where p is a positive integer
I
real number
-k
00
(k+ p)(k + p 1)
00
I=X
e.
I
where a is a positive integer
00
ә
К»
ak +1)(ak + a + 1)'
f. If the series
[=\
I
Vk+3
bconver
00
T
Vk1
k 1
g. Viewed as a
00
67.
00
9
+2k
k25k + 4
9
takes on all
=1
69. (tan(k 1) - tan k)
(- w
00
88-89. Binary nun
00
k-1
base-10 or decimal
k 1
numbers internally
Evaluating an infinite series two ways Evaluate the series
0's and l's. For thi
I
2k 2k+1
00
I+yC
two ways
form 0.b,b,b3
The base-10 repre
a. Use a telescoping series argument.
b. Use a geometric series argument with Theorem 10.8.
71. Evaluating
infinite series
an infinite series two ways Evaluate the series
88. Verify that th

+
Ms M
TAU
456456456.
sin
125=1.252525.
5.12838383.
4 Telescoping series For the following telescoping series, find a
e lim S to obtain the value of the series or state that the series
53. 0.00952
0.00952952.
In ((k 1)k
6
(In k) In (k +
E8275
k 2
for the nth term of the sequence of partial sums S. Then
o0
81. 20
hat
I
k + 1
I
k +2
C0
00
I
83.
9
I ¥
k + 3
+2
57.
I
0
85.
(L(9)
.)
25k215k 4
00
10
87. Explain why or
(4k 3) (4k 1)
ments are true a
= 1
4k2+32k + 63
30*
E-F!
k + 1
61. In
00
00
a. If a con
I
k 1
k-1
( 3k + 1)(3k + 4)
00
b. If a, div
k 1
0=}
2
00
(2k-1)(2k 1)
c. If a co
E 7
where p is a positive integer
d. If p di
I
CC
real numb
. (I + d+ y)(d +
00
[=}
e.
I
where a is a positive integer
#2 ak + 1)(ak + a + 1)'
1=\
f. If the seri
1
I
Vk+ 1
00
C0
Vk+ 3
k 1
(
g. Viewed:
O0
9
k2+5k + 4
99
K+2k
takes on
3
= \
00
69. (tan(k + 1) - tan k)
88-89. Binary
k 1
base-10 or deci
Evaluating an infinite series two ways Evaluate the series
numbers interne
0's and l's. For
I
two ways.
form 0.b,b,b3
The base-10 re
2 2k+1
.
=>
2 Use a telescoping series argument.
b. Use a geometric series argument with Theorem 10.8.
infinite series

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