Answer
The series converges if $-1 \lt x \lt 5$ and the sum is $\frac{3}{5-x}$
Work Step by Step
Given:
$\Sigma^{\infty}_{n=0} \frac{(x-2)^{n}}{3^{n}}$
The given series can be re-written as
$\Sigma^{\infty}_{n=0} (\frac{x-2}{3})^{n}$
Here,
$a=1$ and $r=(\frac{x-2}{3})$
Also, $|r| \lt 1$
$|\frac{x-2}{3}| \lt 1$
$|x-2| \lt 3$
$-3 \lt x-2 \lt 3$
$-1 \lt x \lt 5$
Therefore, the series is convergent.
The sum can be calculated as follows:
$\Sigma^{\infty}_{n=1} \frac{(x-2)^{n}}{3^{n}} = \frac{a}{1-r}$
$=\frac{1}{1-(\frac{x-2}{3})}$
$=\frac{1}{(\frac{3-x+2}{3})}$
$=\frac{3}{5-x}$
