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How to sum $2^{2m}$ where $m$ varies from $0$ to $n$?

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2 Answers

up vote 3 down vote accepted

Note that you can get: $2^{2m}=4^m$, now

$$\displaystyle\sum_{m=0}^{n} 2^{2m} = \displaystyle\sum_{m=0}^{n} 4^{m}$$

The last is easy, (is geometric with $r=4$), so

$$\sum_{m=0}^{n} 2^{2m} = \sum_{m=0}^{n} 4^{m} = \frac{4-4^{n+1}}{1-4}+1=\frac{4^{n+1}-4}{3}+1=\frac{4^{n+1}-1}3.$$

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If you really mean $n$ varies from $0$ to $n$ then the answer is $\sum _{n=0}^n 2^{2 m}=2^{2 m} (1 + n)$

Otherwise if $m$ goes from $0$ to $n$ its just a geometric series ( http://mathworld.wolfram.com/GeometricSeries.html )

$\sum _{m=0}^n 2^{2 m}=\frac{1}{3} \left(2^{2 n+2}-1\right)$

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The first sum is nonsense. –  AD. Jan 5 '11 at 17:41
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Actually he said that $n$ varies from $0$ to $n$. And yes basically its nonsense, you can however interpret it as a sum without condition that just loops n+1 times, common CAS-Systems accept it you just have to make clear that the n=0 and n which limits the sum are NOT the same. They just have the same name. –  Listing Jan 5 '11 at 17:43
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