Answer
$R=\frac{1}{2}$ ; the interval of convergence is $(-\frac{1}{2},\frac{1}{2})$
Work Step by Step
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{2^{(n+1)}(n+1)^2x^{n+1}}{2^nn^2x^n}|$
$=2|x|$
Hence, $R=\frac{1}{2}$ ; the interval of convergence is $(-\frac{1}{2},\frac{1}{2})$
