Answer
$R=2$
$I=(-2,2)$
Work Step by Step
Given $$\sum_{n=1}^{\infty} (-1)^n \frac{n^2x^n}{2^n} $$
Since $a_n=(-1)^n \dfrac{n^2x^n}{2^n} $, $a_{n+1}=(-1)^{n+1} \dfrac{(n+1)^2x^{n+1}}{2^{n+1}} $, then
\begin{align*}
\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\to\infty}\left|(-1)^{n+1} \dfrac{(n+1)^2x^{n+1}}{2^{n+1}} \dfrac{2^n}{(-1)^nn^2x^n}\right|\\
&=\lim_{n\to\infty}\left| \dfrac{(n+1)^2x }{2n^2 } \right| \\
&=\frac{|x|}{2 }\lim_{n\to\infty} \left( 1+\dfrac{1 }{ n}\right)^2\\
&= \frac{|x|}{2 }
\end{align*}
By the Ratio Test, the series $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{n^2x^n}{2^n} $ converges when $ \frac{|x|}{2 }\lt1\ \Rightarrow (-2,2)$
