Answer
$\displaystyle \sum_{i=1}^{6}(\frac{1}{3})^{i-1}=\frac{364}{243}$
Work Step by Step
$\displaystyle \sum_{i=1}^{6}(\frac{1}{3})^{i-1}=(\frac{1}{3})^{0}+(\frac{1}{3})^{1}+..+(\frac{1}{3})^{5}$
$a_1=(\frac{1}{3})^0=1$
$r=\frac{a_2}{a_1}=\frac{(\frac{1}{3})^{1}}{(\frac{1}{3})^{0}}=\frac{1}{3}$
$S_n=a_1(\frac{1-r^n}{1-r})$
$S_6=1(\frac{1-(\frac{1}{3})^6}{1-\frac{1}{3}})=\frac{1-\frac{1}{729}}{\frac{2}{3}}=\frac{\frac{728}{729}}{\frac{2}{3}}=\frac{364}{243}$
