I am having trouble trying to solve this summation (with fraction)
$$f(n) = \sum\limits_{i=1}^{n} \frac{-5}{6^i}$$
How do I solve this?
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For $0\leq m\leq n$ and $r\not= 1$,
$$\sum_{k = m}^{m}ar^{k} = \frac{a(r^m - r^{n+1})}{1-r}.$$
Hence, $$\sum_{k=1}^{n}\frac{-5}{6^k} = \frac{(-5)(\frac{1}{6}-(\frac{1}{6})^{n+1})}{1-\frac{1}{6}}$$ $$ = (-6)\left(\frac{1}{6}-\left(\frac{1}{6}\right)^{n+1}\right)$$ $$ = (-1)\left(1-\left(\frac{1}{6}\right)^{n}\right)$$ $$ = \frac{1}{6^{n}} - 1$$
prob_noob
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For $0\leq m\leq n$ and $r\neq 1$ $$ \sum_{k=m}^nr^k=\frac{r^m-r^{n+1}}{1-r}. $$ See geometric progression.
Now note $$ \sum_{k=1}^n\frac{-5}{6^k}=-5\sum_{k=1}^n\frac{1}{6^k}=-5\sum_{k=1}^n\left(\frac{1}{6}\right)^k. $$
Julien
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But how do I "move" 6^i up and get rid of the fraction? Thanks! – monsmanr Mar 20 '13 at 02:51
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@monsmanr See the edit. – Julien Mar 20 '13 at 02:53