Answer
$S_n=2[1-( \dfrac{2}{3})^n]$
Work Step by Step
The sum of the First $n$ Terms of a Geometric Sequence is given by:
$S_{n}=\displaystyle\sum_{k=1}^n a_1r^{k-1}=a_{1} (\dfrac{1-r^{n}}{1-r}) ; \ r\neq 0,1$
We are given: $a_{1}=\dfrac{2}{3} ; \ r= \dfrac{2}{3}$
Thus, we can write the sum as:
$\displaystyle\sum_{k=1}^n (\dfrac{2}{3})^{k}=\displaystyle\sum_{k=1}^n (\dfrac{2}{3}) (\dfrac{2}{3})^{k-1}$
Now, $S_n= \dfrac{2}{3} \ [\dfrac{1-(\dfrac{2}{3} )^{n}}{1- \dfrac{2}{3} } \ ] \\=\dfrac{2}{3} \ [\dfrac{1-(\dfrac{2}{3} )^{n}}{ \dfrac{1}{3} } \ ] $
Therefore, $S_n=2[1-( \dfrac{2}{3})^n]$
