## Algebra 2 (1st Edition)

$\approx 2059.94$
Here, we have $\sum_{i=1}^{12} 8(3/2)^{i-1}$ We know that $S_{n}=a_1(\dfrac{1-r^{n}}{1-r})$ Common ratio $r=\dfrac{3}{2}$ and $a_1= 8$ Now, $S_{12}=8[\dfrac{1-(3/2)^{12}}{1-(3/2)}]$ Hence, $S_{12}=8 \times [\dfrac{1-(3/2)^{12}}{1-(3/2)}] \approx 2059.94$