Answer
The series is convergent if $\frac{19}{4} \lt x \lt \frac{21}{4}$ and the sum is $\frac{1}{4x-19}$.
Work Step by Step
Given: $\Sigma^{\infty}_{n=0} (-4)^{n}(x-5)^{n} $
$\Sigma^{\infty}_{n=0} (-4)^{n}(x-5)^{n} = \Sigma^{\infty}_{n=0}(-4x+20)^{n}$
Here, $a=1$ and $r=(-4x+20)^{n}$
Also, $|r| \lt 1$
$|-4x + 20| \lt 1$
$-1 \lt -4x +20 \lt 1$
$-21 \lt -4x \lt -19$
$\frac{19}{4} \lt x \lt \frac{21}{4}$
The series is convergent if $\frac{19}{4} \lt x \lt \frac{21}{4}$
The sum can be calculated as follows:
$\Sigma^{\infty}_{n=0} (-4)^{n} (x-5)^{n} = \frac{a}{1-r}$
$=\frac{1}{1-(-4x+20)}$
$=\frac{1}{4x-19}$
