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

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