Sum of series $2^{2m}$

Question

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

Answer 1

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

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

Answer 2

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 )

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