Answer
$\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(n+1)(n+2)x^{n}}{2^{(n+4)}}$
and
$R=2$
Work Step by Step
$\frac{1}{(2+x)^{3}}=(2+x)^{-3}=2^{-3}(1+\frac{x}{2})^{-3}=\frac{1}{8}(1+\frac{x}{2})^{-3}$
$=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(n+1)(n+2)x^{n}}{2^{(n+4)}}$
$$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(n+2)(n+3)x^{n+1}}{2^{(n+5)}}}{\frac{(n+1)(n+2)x^{n}}{2^{(n+4)}}}|$$
$=\lim\limits_{n \to\infty}|\frac{(n+3)x}{2(n+2)}|$
$=|\frac{x}{2}|$
The series will converge when $|\frac{x}{2}|\lt 1$, or $|x|\lt 2$ so $R=2$.
