## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 578: 42

#### Answer

$$\sum_{n=0}^{\infty} \frac{(x+2)^{n}}{3^{n+1}}$$ $|x+2|\lt 3$ $$\sum_{n=0}^{\infty}(-1)^{n+1}(x-2)^{n}$$ $|x-2|\lt1$

#### Work Step by Step

Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$ For $c=-2$, by using (1), we get \begin{align*} \frac{1}{1-x}&=\frac{1}{3-(x+2)}\\ & =\frac{\frac{1}{3}}{1-\frac{(x+2)}{3}}\\ & =\frac{1}{3} \frac{1}{1-\frac{(x+2)}{3}}\\ &=\frac{1}{3} \sum_{n=0}^{\infty}\left( \frac{(x+2)}{3}\right)^n\\ &= \sum_{n=0}^{\infty} \frac{(x+2)^{n}}{3^{n+1}} \end{align*} Such that $|x+2|\lt 3$ For $c= 2$, by using (1), we get \begin{align*} \frac{1}{1-x}&=\frac{1}{-1-(x-2)}\\ & =\frac{-1}{1+(x-2)}\\ & =\frac{-1}{1-(-(x-2))}\\ &=(-1) \sum_{n=0}^{\infty}(-1(x-2))^{n}\\ &= \sum_{n=0}^{\infty}(-1)^{n+1}(x-2)^{n} \end{align*} Such that $|x-2|\lt1$

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