#### Answer

$131070$

#### Work Step by Step

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, $=131070$