(- 1)k + 1 1 0k- 1 kx2k-1 = 0 +x -20x3 300x5 -40000x7 + ... in its expanded The power series for g(x) is k 0 form. Drop the zero term and re-index the series so that k 0 corresponds to the term x 0o g(x)>(-1)k10K (k+ 1)x2k+1 k 0

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.3: Geometric Sequences
Problem 44E
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I am working with power series and this answer explanation shown in the screen shot does not make sense to me. Why and are they reindexing and how to they get that resulting answer?

(- 1)k + 1 1 0k- 1 kx2k-1 = 0 +x -20x3 300x5 -40000x7 + ...
in its expanded
The power series for g(x) is
k 0
form. Drop the zero term and re-index the series so that k 0 corresponds to the term x
0o
g(x)>(-1)k10K (k+ 1)x2k+1
k 0
Transcribed Image Text:(- 1)k + 1 1 0k- 1 kx2k-1 = 0 +x -20x3 300x5 -40000x7 + ... in its expanded The power series for g(x) is k 0 form. Drop the zero term and re-index the series so that k 0 corresponds to the term x 0o g(x)>(-1)k10K (k+ 1)x2k+1 k 0
Expert Solution
Step 1

Note that the variable k  lies inside the summation as a product terms. So, when k = 0, the first term of the series becomes zero.

Σ-)0 ke- (-1)** 10*' (0) >20 Σ-1)"
k+1
= (-1)'10-1(0)x20)-1
2k-1
Σ-1)"10*" kκ.
k-0
k-1
k+1
-1x10' x 0x x1 +1)10-1 κ
2k-1
k1
k+1
0(-1) 10*-1kx2*-1
k1
k+1
-Σ-1)104" kλ-
Ε
k-1
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