Answer
Interval of Convergence : $-4 \lt x \lt 2$
and $\Sigma_{n=0}^{\infty} \dfrac{(x+1)^{2n}}{9^n}=\dfrac{9}{8-2x-x^2}$
Work Step by Step
We have $\Sigma_{n=0}^{\infty} \dfrac{(x+1)^{2n}}{9^n}= \Sigma_{n=0}^{\infty} [\dfrac{x^2+2x+1}{9}]^n=\Sigma_{n=0}^{\infty} (f(x))^n$
The series will converge when $|f(x)|=|\dfrac{x^2+2x+1}{9}| \lt 1$
So, $|\dfrac{(x+1)^2}{9}| \lt 1 \implies -4 \lt x \lt 2$
Now, the sum of a given geometric series is:
$S=\dfrac{a}{1-r}=\dfrac{1}{1-\dfrac{x^2+2x+1}{9}}=\dfrac{9}{8-2x-x^2}$
Thus, the Interval of Convergence is:
$-4 \lt x \lt 2$
and $\Sigma_{n=0}^{\infty} \dfrac{(x+1)^{2n}}{9^n}=\dfrac{9}{8-2x-x^2}$
