Answer
$\displaystyle \frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{3^{n}}$
Work Step by Step
We see that there are $n$ terms, as the index $k$ changes from $0$ to $n$. The index $k$ indicates how the terms are formed.
We write out the sum for the terms as follows:
$\displaystyle \sum_{k=0}^{n-1}\frac{1}{3^{k+1}}=\frac{1}{3^{(0+1)}}+\frac{1}{3^{(1+1)}}+\frac{1}{3^{(2+1)}}+...+\frac{1}{3^{(n-1+1)}} \\ =\displaystyle \frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{3^{n}}$
