Answer
$\frac{4}{3}+\frac{4x}{9}+\frac{4x^2}{27}+...\frac{4}{3}(\frac{x}{3})^n+...$
Work Step by Step
We are given $f(x)=\frac{4}{3-x}=\frac{\frac{4}{3}}{1-\frac{x}{3}}$
for $r$ in $(-3,3)$
The Taylor series for $\frac{1}{1-x}$ is
$1+x+x^2+x^3+...+x^n+...$
The Taylor series for $f(x)=\frac{\frac{4}{3}}{1-\frac{x}{3})}$ is
$f(x)=\frac{4}{3}.1+\frac{4}{3}(\frac{x}{3})+\frac{4}{3}(\frac{x}{3})^2+\frac{4}{3}(\frac{x}{3})^3...+\frac{4}{3}(\frac{x}{3})^n+...$
$=\frac{4}{3}+\frac{4x}{9}+\frac{4x^2}{27}+...\frac{4}{3}(\frac{x}{3})^n+...$
