Answer
$1+\displaystyle \frac{3}{2}+\frac{9}{4}+\frac{27}{8}+..+(\frac{3}{2})^{n}$
Work Step by Step
We see that there are $n+1$ 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 $n$ terms as follows:
$\displaystyle \sum_{k=0}^{n} (\frac{3}{2})^{k}=(\frac{3}{2})^{0}+(\frac{3}{2})^{1}+(\frac{3}{2})^{2}+...+\left(\frac{3}{2}\right)^{n} \\=1+\displaystyle \frac{3}{2}+\frac{9}{4}+\frac{27}{8}+...+(\frac{3}{2})^{n}$
