Answer
For $r$ in $(-2,2)$
The Taylor series for $f(x)=\frac{3x^3}{2-x}$ is
$\frac{3x^3}{2}+\frac{3x^4}{4}+\frac{3x^5}{8}+\frac{3x^6}{16}...+\frac{3(x^{n+3})}{2^{n+1}}+...$
Work Step by Step
We are given $f(x)=\frac{3x^3}{2-x}=\frac{\frac{3x^3}{2}}{1-\frac{x}{2}}$
for $r$ in $(-2,2)$
The Taylor series for $\frac{1}{1-x}$ is
$1+x+x^2+x^3+...+x^n+...$
The Taylor series for $f(x)=\frac{3x^3}{2-x}$ is
$f(x)=\frac{3x^3}{2}.1+\frac{3x^3}{2}(\frac{x}{2})+\frac{3x^3}{2}(\frac{x}{2})^2+\frac{3x^3}{2}(\frac{x}{2})^3...+\frac{3x^3}{2}(\frac{x}{2})^n+...$
$=\frac{3x^3}{2}+\frac{3x^4}{4}+\frac{3x^5}{8}+\frac{3x^6}{16}...+\frac{3(x^{n+3})}{2^{n+1}}+...$
