Answer
$1+\displaystyle \frac{3}{2}+\frac{9}{4}+\frac{27}{8}+\cdots+\left(\frac{3}{2}\right)^{n}$
Work Step by Step
There are n+1 terms, as the index k changes from 0 to n.
The index k dictates how the terms are formed:
$\displaystyle \sum_{k=0}^{n}\left(\frac{3}{2}\right)^{k}=\left(\frac{3}{2}\right)^{0}+\left(\frac{3}{2}\right)^{1}+\left(\frac{3}{2}\right)^{2}+...+\left(\frac{3}{2}\right)^{n}$
$=1+\displaystyle \frac{3}{2}+\frac{9}{4}+\frac{27}{8}+\cdots+\left(\frac{3}{2}\right)^{n}$
