# How do you write the formula for the nth term given 8, -16, 32, -64,...?

${\left(- 1\right)}^{n - 1} {2}^{n + 2}$
It is a geometric series, with first term 8 and common ratio -2, hence the nth term would be $a {r}^{n - 1}$
= $8 {\left(- 2\right)}^{n - 1}$
= ${\left(- 1\right)}^{n - 1} \left(8\right) {2}^{n - 1}$. This can also be written as ${\left(- 1\right)}^{n - 1} {2}^{n + 2}$