Answer
$R=\frac{1}{2}$
and
Interval of convergence:$(-\frac{1}{2},\frac{1}{2})$
Work Step by Step
Given: $a_n=2^nn^2x^n$ and $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{2^{(n+1)}(n+1)^2x^{n+1}}{2^nn^2x^n}|=2|x|$
Hence,
$R=\frac{1}{2}$
and
Interval of convergence:$(-\frac{1}{2},\frac{1}{2})$
