Answer
$S_n=5[1-( \dfrac{3}{5})^n]$
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: $a_1=2$ and $r=\frac{3}{5}$
We can write the sum as:
$S_n=\displaystyle\sum_{k=1}^n (2) (\dfrac{3}{5})^{n-1}$
Now, $S_n= (2) \ [\dfrac{1-(\dfrac{3}{5} )^{n}}{1- \dfrac{3}{5} } \ ] \\=(2) \ [\dfrac{1-(\dfrac{3}{5} )^{n}}{ \dfrac{2}{5} } \ ] $
Therefore, $S_n=5[1-( \dfrac{3}{5})^n]$
