Let suppose
$S=\sum_{j=2}^{8}(-3)^{j}$
$S=(-3)^{2}+(-3)^{3}+(-3)^{4}+(-3)^{5}+(-3)^{6}+(-3)^{7}+(-3)^{8}$
$S=(3)^{2}-(3)^{3}+(3)^{4}-(3)^{5}+(3)^{6}-(3)^{7}+(3)^{8}$
$S=(3)^{2}+(3)^{4}+(3)^{6}+(3)^{8}-(3)^{3}-(3)^{5}-(3)^{7}$
$S=(3)^{2}+(3)^{4}+(3)^{6}+(3)^{8}-[(3)^{3}+(3)^{5}+(3)^{7}]$
Let say $S_{1}=(3)^{2}+(3)^{4}+(3)^{6}+(3)^{8}$ and $S_{2}=(3)^{3}+(3)^{5}+(3)^{7}$
Now $S=S_{1}-S_{2}$-----------------$>(1)$
First find the value of $S_{1}=(3)^{2}+(3)^{4}+(3)^{6}+(3)^{8}$ this is the $GP$ with $a=(3)^{2},r=(3)^{2},n=4$
$S_{1}=\frac{a(r^{n}-1)}{(r-1)};r>1$
$S_{1}=\frac{(3)^{2}[((3)^{2})^{4}-1]}{[(3)^{2}-1]}$
$S_{1}=\frac{(3)^{2}.[(3)^{8}-1]}{[(3)^{2}-1]}$
$S_{1}=\frac{(3)^{10}-(3)^{2}}{[(3)^{2}-1]}$
and for $S_{2}=(3)^{3}+(3)^{5}+(3)^{7}$ this is also a $GP$ with $a=(3)^{3},r=(3)^{2},n=3$
$S_{2}=\frac{a(r^{n}-1)}{(r-1)};r>1$
$S_{2}=\frac{(3)^{3}[((3)^{2})^{3}-1]}{[(3)^{2}-1]}$
$S_{2}=\frac{(3)^{3}.[(3)^{6}-1]}{[(3)^{2}-1]}$
$S_{2}=\frac{(3)^{9}-(3)^{3}}{[(3)^{2}-1]}$
Put the value of $S_{1}$ $and $ $S_{2}$ in equation $(1)$
Now we got $S=\frac{(3)^{10}-(3)^{2}}{[(3)^{2}-1]}-\frac{(3)^{9}-(3)^{3}}{[(3)^{2}-1]}$
$S=4,923$