How to sum $2^{2m}$ where $m$ varies from $0$ to $n$?
Sum of series $2^{2m}$
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$\begingroup$
summation
geometric-progressions
2 Answers
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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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2Actually 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. – 2011-01-05