Answer
$$\sum_{n=0}^{\infty}\frac{(-1)^n3^nx^n}{4^{n+1}}$$
$|x|\lt4/3$
Work Step by Step
Given $$ f(x)=\frac{1}{4+3x}$$
Since
$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt1\tag{1}$$
By using (1), we get
\begin{align*}
\frac{1}{4+3x}&=\frac{1}{4(1+(3x/4))}\\
&=\frac{1/4}{(1-(-3x/4))}\\
&=\frac{1}{4} \sum_{n=0}^{\infty}( -3x/4)^{n}\\
&= \frac{1}{4} \sum_{n=0}^{\infty}\frac{(-3)^nx^n}{4^n}\\
&= \sum_{n=0}^{\infty}\frac{(-1)^n3^nx^n}{4^{n+1}}\\
\end{align*}
Such that $ |3x/4|\lt 1$, or $|x|\lt4/3$
