# Formula for a geometric series

Helping my brother with his math homework and I am hung up on this one.

Find a formula for $a_n$ for the geometric sequence:
$a_1=2$, $a_{k+1}=-3a_k$

If anyone here can help that would be great. Thanks. My thought would be that I just need to turn the formula above into $a_k = ____ a_{k-1}$

$a_n=2\cdot (-3)^{n-1}$, or $\,(-1)^{n-1}\, 2\cdot 3^{n-1}$ , if the sequence indices begin at $n=1$.
$$a_n = (-1)^{n+1} (2)( 3^{n-1})$$
• May I suggest $(-1)^{n+1}$ is the same as $(-1)^{n-1}$? – Bernard Mar 18 '15 at 0:40
I think you are searching for $a_n = 2\cdot(-3)^n$