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I am given the following geometric series and am asked to find the sum.

$$\sum_{n=1}^{\infty} \left(\frac{12}{(-5)^n}\right)$$

I know that I somehow need to get this in the form $\sum_{n=1}^{\infty}ar^{n-1}$, where $a$ is the first term and $r$ is the ratio, but the best I could come up with is the following:

$$\sum_{n=1}^{\infty} \left(12(-5)^{-n}\right)$$

However, It needs to be in the following form: $$\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}$$

I am not sure what I am missing here... can someone point me in the right direction?

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$$12\;(-5)^{-n}=12\;\left(-\frac{1}{5}\right)^n=-\frac{12}{5}\left(-\frac{1}{5}\right)^{n-1}$$

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  • $\begingroup$ Right, but don't I need $n-1$, not $n$ in the exponent for the equality I mentioned above to work? $\endgroup$ – dtg Apr 12 '13 at 6:10
  • $\begingroup$ @Dylan There you go. $\endgroup$ – Julien Apr 12 '13 at 6:10
  • $\begingroup$ Urg... factoring... thank you. $\endgroup$ – dtg Apr 12 '13 at 6:12

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