Answer
$S_6=1330$
Work Step by Step
We need to find:
$\sum_{k=1}^{6}64(\frac{3}{2})^{k-1}$
We see that this is a geometric sequence with $a=64$ and $r=\frac{3}{2}$.
We know the partial sum of a geometric sequence is:
$S_n=a_1\frac{1-r^n}{1-r}$
$S_6=64\frac{1-(\frac{3}{2})^{6}}{1-(\frac{3}{2})}=1330$