Answer
$\frac{-3}{4}-\frac{3x}{16}-\frac{3x^2}{64}-\frac{3x^3}{256}-...-\frac{3x^n}{4^{n+1}}-...$
Work Step by Step
We are given $f(x)=\frac{-3}{4-x}=\frac{\frac{-3}{4}}{1-\frac{x}{4}}=\frac{-3}{4}\frac{1}{1-\frac{x}{4}}$
with $c=\frac{-3}{4}$ for $r$ in $(-4,4)$
The Taylor series for $f(x)=\frac{-3}{4-x}$ is
$f(x)=\frac{-3}{4}.1+\frac{-3}{4}(\frac{x}{4})+\frac{-3}{4}(\frac{x}{4})^{2}+\frac{-3}{4}(\frac{x}{4})^{3}...+\frac{-3}{4}(\frac{x}{4})^n+...$
$=\frac{-3}{4}-\frac{3x}{16}-\frac{3x^2}{64}-\frac{3x^3}{256}-...-\frac{3x^n}{4^{n+1}}-...$
