Answer
geometric, sum=$9831$
Work Step by Step
A geometric sequence has a common ratio
$a_{n}=ar^{n-1}$, with nth partial sum:
$S_{n}=a\displaystyle \cdot\frac{1-r^{n}}{1-r}$
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The sum has 7 terms (for n=0,1,2,...6).
We can rewrite it as
$\displaystyle \sum_{n=1}^{7}3(-4)^{n-1}$
The terms make up a geometric sequence, $a=3, r=-4$
$a_{n}=3(-4)^{n-1}$
The sum equals
$S_{n}=a\displaystyle \cdot\frac{1-r^{n}}{1-r}=3\cdot\frac{1-(-4)^{7}}{1-(-4)}$
$=\displaystyle \frac{3(1+4^{7})}{5}=9831$
