Answer
$1+\frac{3}{2}+\frac{9}{4}+...+\frac{3^n}{2^n}$
Work Step by Step
$\sum_{k=0}^{n} \left(\frac{3}{2}\right)^k=\left(\frac{3}{2}\right)^0+\left(\frac{3}{2}\right)^1+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^n=1+\frac{3}{2}+\frac{9}{4}+...+\frac{3^n}{2^n}$
