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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 $a_k = ____ a_{k-1}$

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$a_n=2\cdot (-3)^{n-1}$, or $\,(-1)^{n-1}\, 2\cdot 3^{n-1}$ , if the sequence indices begin at $n=1$.

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$$ a_n = (-1)^{n+1} (2)( 3^{n-1}) $$

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  • $\begingroup$ May I suggest $ (-1)^{n+1}$ is the same as $ (-1)^{n-1}$? $\endgroup$ – Bernard Mar 18 '15 at 0:40
  • $\begingroup$ Yes, you are right. Thanks you ! $\endgroup$ – alex14204 Mar 18 '15 at 0:45
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I think you are searching for $a_n = 2\cdot(-3)^n$

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