Answer
$$\sum_{n=0}^{\infty}(-1)^{n+1} \frac{(x-4)^{n}}{3^{n+1}}$$
$ |x-4|\lt 3$
Work Step by Step
Given
$$ \frac{1}{1-x}=\frac{1}{-3-(x-4)}=\frac{-\frac{1}{3}}{1+\left(\frac{x-4}{3}\right)}$$
Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$
By using (1), we get
\begin{align*}
\frac{1}{1-x}&=\frac{-1}{3}\frac{1}{1+\left(\frac{x-4}{3}\right)}\\
&=\frac{-1}{3}\sum_{n=0}^{\infty}\left(-\frac{x-4}{3}\right)^{n}\\
&=\frac{-1}{3}\sum_{n=0}^{\infty}(-1)^{n} \frac{(x-4)^{n}}{3^{n}}\\
&= \sum_{n=0}^{\infty}(-1)^{n+1} \frac{(x-4)^{n}}{3^{n+1}}
\end{align*} Such that $ |x-4|\lt 3$
