Answer
$\frac{x^{2}}{4}+\frac{x^{3}}{16}+\frac{x^{4}}{64}+...+\frac{x^{2+n}}{4^{n+1}}+...$
Work Step by Step
We are given $f(x)=\frac{x^{2}}{4-x}=\frac{\frac{x^{2}}{4}}{1-\frac{x}{4}}$
with $c=\frac{x^{2}}{4}$ for $r$ in $(-4,4)$
The Taylor series for $f(x)=\frac{x^{2}}{4-x}$ is
$f(x)=\frac{x^{2}}{4}+\frac{x^{2}}{4}(\frac{x}{4})+\frac{x^{2}}{4}(\frac{x}{4})^{2}+\frac{x^{2}}{4}(\frac{x}{4})^3+...+\frac{x^{2}}{4}(\frac{x}{4})^n+...$
$=\frac{x^{2}}{4}+\frac{x^{3}}{16}+\frac{x^{4}}{64}+...+\frac{x^{2+n}}{4^{n+1}}+...$
