Answer
$a_n=2(-4)^{n-1}$
Work Step by Step
$ 2, -8, 32, -128, 512, -2048, . . .$
$\frac{a_2}{a_1}=\frac{-8}{2}=-4=r$
$ra_2=-4\times-8=32=a_3$
$ra_3=-4\times32=-128=a_4$
and so on. The sequence is geometric with common ratio $r=-4$
The $n^{th}$ term of the sequence is given by
$a_n=2r^{n-1}=2(-4)^{n-1}$
