## Sum of Series with negative exponents [closed]

$3= \sum_{n=1}^{t} \frac{1}{1.08^n}$

I see that it is $3 = 1.08^{-t}(12.5 \times 1.08^t{-12.5})$ (from Wolfram Alpha, but I'm not sure how to get it. I tried solving as a geometric series, I had problems and didn't get the correct answer.

I see that $t\approx3.56592$, which seems like it's correct, but I have no idea where the 12.5 and all that stuff came from. Unfortunately my calculus book doesn't help much, as it is mainly focus on infinite series.

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 Your question would be more suited to NRICH or one of the other sites mentioned in the FAQ mathoverflow.net/faq#whatnot – Yemon Choi Oct 6 2010 at 21:12 You could try math.stackexchange.com, as the FAQ would have told you. – Mariano Suárez-Alvarez Oct 6 2010 at 21:12