Answer
$\sum_{n=0}^{\infty}n(n-1)\frac{x^{n+1}}{2^{n+2}}$,
$R=2$
Work Step by Step
$f(x)=(\frac{x}{2-x})^{3}=\sum_{n=2}^{\infty}n(n-1)\frac{x^{n+1}}{2^{n+2}}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(n+1)((n+1)-1)\frac{x^{n+2}}{2^{n+3}}}{n(n-1)\frac{x^{n+1}}{2^{n+2}}}|$
$=|\frac{x}{2}|\lt 1$
The given series converges with $R=2$
