Answer
$$\sum_{n=0}^{\infty}(-1)^{n+1}(x-5)^{n}$$
$ |x-5|<1 $
Work Step by Step
Given $$\frac{1}{4-x}$$
Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$
For $c=5 $, by using (1), we get
\begin{aligned}
\frac{1}{4-x}&=\frac{1}{-1-(x-5)}\\
& =\frac{-1}{1+(x-5)}\\
& =\frac{-1}{1-(-(x-5))}\\
&=(-1) \sum_{n=0}^{\infty}(-1(x-5))^{n}\\
&=\sum_{n=0}^{\infty}(-1)^{n+1}(x-5)^{n}
\end{aligned}
Such that $ |x-5|<1 $
