# Simplifying this sigma notation [closed]

Is there any way I can simplify this sigma notation?

\begin{align*} \sum_{k=1}^m(5^k) \end{align*}

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## closed as off-topic by alexqwx, Gina, Tomás, leo, StahlAug 6 '14 at 20:27

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– lab bhattacharjee Aug 6 '14 at 12:54
This is simply a geometric sum: $$\sum_{k=1}^m x^k = x\cdot \frac{1-x^{m}}{1-x}$$ – Joel Aug 6 '14 at 12:58

Yes, there is.

Let $S = \displaystyle\sum_{k=1}^{m} 5^{k}$ be the sum you're trying to simplify, then you'll have $$5S=\color{red}{5^2+5^3+\cdots+5^m}+5^{m+1}$$ $$S=5^1+\color{red}{5^2+\cdots+5^{m-1}+5^m}$$ Then, substracting $S$ from $5S$ gives you $$4S=5^{m+1}-5^1\iff S=\frac{5^{m+1}-5}{4}.$$

In general, for $a\not =1$, you'll get the following using the above method : $$\sum_{k=1}^{m}a^k=\frac{a^{m+1}-a}{a-1}.$$ Note that if $a=1$, then $\sum_{k=1}^{m}1^k=m$ trivially.

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@Ogen: I hope this helps:) – mathlove Aug 6 '14 at 18:23
@EvanW: Thanks:) better now. – mathlove Aug 6 '14 at 18:53
thanks, your answer is a good demonstration. – Evan W Aug 6 '14 at 18:55
@mathlove I believe your last statement is incorrect; if $a = 1$, then $a - 1 = 0$, so the sum would be $\infty$; this is because $aS - S = (a - 1)S = 0\times S = 0$, so then you cannot actually solve for $S$. – feralin Aug 6 '14 at 19:16
@feralin: I edited it. That's what I meant. – mathlove Aug 6 '14 at 19:18

Using the formula $$\sum_{k=0}^{N} x^k=\frac{x^{N+1}-1}{x-1}$$ we have:

$$\sum_{k=1}^m 5^k=\sum_{k=1}^m 5^k+5^0-5^0=\sum_{k=0}^m 5^k-1=\frac{5^{m+1}-1}{5-1}-1=\frac{5^{m+1}-1}{4}-1=\frac{5^{m+1}-1-4}{4}=\frac{5^{m+1}-5}{4}$$

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$5+5^2+5^3+...+5^m=5\large\frac{(5^m-1)}{(5-1)}$

It is a GP with first term=$5$ and common ration=$5$

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