1- (2/5) 31 k=0 10.3.19 1-(-2/7)" 1-(-2/7) a. The nth partial sum is S =a So the sum of the series is S = lim Sn = T 00 (1 - 0) = 9.

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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How did we get a (circled in green, says 7/9). How did we get this?

k
1
1- (2/5)
b. S =>
%3D
k=0
10.3.19
1-(-2/7)"
1-(-2/7)
7
a. The nth partial sum is Sn
So the sum of the series
= a
%3D
1-r
7
is S = lim Sn
(1 – 0) =
%3D
%3D
%3D
9.
k
1
7.
|
25
1.
Transcribed Image Text:k 1 1- (2/5) b. S => %3D k=0 10.3.19 1-(-2/7)" 1-(-2/7) 7 a. The nth partial sum is Sn So the sum of the series = a %3D 1-r 7 is S = lim Sn (1 – 0) = %3D %3D %3D 9. k 1 7. | 25 1.
2
19.
7.
K3D0
Transcribed Image Text:2 19. 7. K3D0
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