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$
