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