Answer
$\frac{1}{3}+\frac{1}{9}+...+\frac{1}{3^n}$
Work Step by Step
$\sum_{k=0}^{n-1} \left(\frac{1}{3^{k+1}}\right)=\left(\frac{1}{3^{0+1}}\right)+\left(\frac{1}{3^{1+1}}\right)+...+\left(\frac{1}{3^{(n-1)+1}}\right)=\frac{1}{3}+\frac{1}{9}+...+\frac{1}{3^n}$
