Answer
Interval of Convergence: $-1 \lt x \lt 3$
and $\Sigma_{n=0}^{\infty} \dfrac{(x-1)^{2n}}{4^n}=\dfrac{4}{3+2x-x^2}$
Work Step by Step
We have $\Sigma_{n=0}^{\infty} \dfrac{(x-1)^{2n}}{4^n}= \Sigma_{n=0}^{\infty} [\dfrac{x^2-2x+1}{4}]^n=\Sigma_{n=0}^{\infty} (f(x))^n$
The series will converge when $|f(x)|=|\dfrac{x^2-2x+1}{4}| \lt 1$
So, $|\dfrac{(x-1)^2}{4}| \lt 1 \implies -1 \lt x \lt 3$
Now, the sum of a given geometric series is:
$S=\dfrac{a}{1-r}=\dfrac{1}{1-\dfrac{x^2-2x+1}{4}}=\dfrac{4}{3+2x-x^2}$
Thus, the Interval of Convergence is:
$-1 \lt x \lt 3$
and $\Sigma_{n=0}^{\infty} \dfrac{(x-1)^{2n}}{4^n}=\dfrac{4}{3+2x-x^2}$