Answer
$\displaystyle \sum_{n=1}^{10}(\frac{3}{2})^{n-1}=\frac{58025}{512}$
Work Step by Step
$\displaystyle \sum_{n=1}^{10}(\frac{3}{2})^{n-1}=(\frac{3}{2})^0+(\frac{3}{2})^1+...+(\frac{3}{2})^9$
$a_1=(\frac{3}{2})^0=1$
$r=\frac{a_2}{a_1}=\frac{(\frac{3}{2})^1}{(\frac{3}{2})^0}=\frac{3}{2}$
$S_{10}=a_1(\frac{1-r^n}{1-r})=1(\frac{1-(\frac{3}{2})^{10}}{1-(\frac{3}{2})})=\frac{1-(\frac{59049}{1024})}{-\frac{1}{2}}=\frac{-(\frac{58025}{1024})}{-\frac{1}{2}}=\frac{58025}{512}$
