Answer
$C$
Work Step by Step
Here, we have $S_9=\sum_{i=1}^9 2(3)^{i-1}$
We know that $S_{n}=a_1(\dfrac{1-r^{n}}{1-r})$
Here, $a_1=2(3)^{1-1}=2$ and $r=3$
Now, $S_9=2[\dfrac{1-(3)^{9}}{1-3}]$
or, $=\dfrac{2 \times (-19682)}{-2}$
or, $S_9=19,682$
Hence, our answer is $C$.