Answer
$2(3^{n}-1)$
Work Step by Step
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, $a_{1}=4, r=3$,
$S_{n}=4\displaystyle \left(\frac{1-3^{n}}{1-3}\right)$
$=4\displaystyle \left(\frac{1-3^{n}}{-2}\right)$
$=2(3^{n}-1)$
