Answer
$\frac{1}{\sqrt 2}(x^{2}-\frac{1}{4}x^{3}+\frac{3}{32}x^{4}-....)$
$=x^2/\sqrt{2}+\sum_{n=1}^{\infty}(-1)^n\frac{1*3*5*...(2n-1)}{n!2^{2n+1/2}}$
$R=2$
Work Step by Step
$\frac {x^{2}}{\sqrt{(2+x)}}=\frac{x^{2}}{\sqrt 2}\frac{1}{\sqrt {1+\frac{x}{2}}}$
$=\frac{x^{2}}{\sqrt 2}(1+\frac{x}{2})^{-1/2}$
$=\frac{1}{\sqrt 2}( x^{2}+(-\frac{1}{2})(\frac{x^{3}}{2})+\frac{(\frac{-1}{2})(\frac{-3}{2})}{2^{2}2!}x^{4}+\frac{(\frac{-1}{2})(\frac{-3}{2})(\frac{-5}{2})}{2^{3}3!}x^{5}+....)$
$=\frac{1}{\sqrt 2}(x^{2}-\frac{1}{4}x^{3}+\frac{3}{32}x^{4}-....)$
$=x^2/\sqrt{2}+\sum_{n=1}^{\infty}(-1)^n\frac{1*3*5*...(2n-1)}{n!2^{2n+1/2}}$
$R=2$
