## Algebra 2 (1st Edition)

$C$
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 $C$.