Answer
$2\displaystyle \left[1-\left(\frac{2}{3}\right)^{n}\right]$
Work Step by Step
$\displaystyle \sum_{k=1}^{n}\left(\frac{2}{3}\right)^{k}=\sum_{k=1}^{n}\frac{2}{3}\left(\frac{2}{3}\right)^{k-1}$
Apply
THEOREM: Sum of the First $n$ Terms of a Geometric Sequence
$S_{n}=\displaystyle \sum_{k=1}^{n}a_{1}r^{k-1}=a_{1}\cdot\frac{1-r^{n}}{1-r},\quad r\neq 0,1$
Here, $a_{1}=\displaystyle \frac{2}{3}, r=\frac{2}{3}$
$S_{n}=a_{1}\displaystyle \left(\frac{1-r^{n}}{1-r}\right)=\frac{2}{3}\cdot\frac{1-\left(\frac{2}{3}\right)^{n}}{1-\frac{2}{3}}$
$=\displaystyle \frac{2}{3}\cdot\frac{1-\left(\frac{2}{3}\right)^{n}}{\frac{1}{3}}$
$=2\displaystyle \left[1-\left(\frac{2}{3}\right)^{n}\right]$
