Answer
$29,523$
Work Step by Step
$\dfrac{a_{n+1}}{a_n}=\dfrac{3^{n+1}}{3^n}=3$
Since the ratio between consecutive terms is constant, the given sequence is geometric.
The first term is $a_{1}=3^{1}=3$.
The common ratio is $r=3$.
The formula for the finite sum of the geometric sequence is:
$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$
$S_{9}=\frac{3(1-3^{9})}{1-3}=29,523$
