## Algebra 2 (1st Edition)

$131070$
Here, we have $\sum_{i=1}^8 6(4)^{i-1}=6\sum_{i=1}^8 (4)^{i-1}$ We know that $S_{n}=a_1(\dfrac{1-r^{n}}{1-r})$ Now, $\sum_{i=1}^8 6(4)^{i-1}=6\sum_{i=1}^8 (4)^{i-1}=1(\dfrac{1-4^{8}}{1-4})$ or, $=6 \times (\dfrac{1-65536}{-3})$ or, $=131070$