Sum with non unit increment Let's consider the sum $$\sum_{i=4t+2} {\binom{m}{i}}$$. 
It's equivalent to the following $\sum_{s}{\binom{m}{4s+2}}$, but i got stuck here.
How to evaluate such kind of sums? For instance, it's not so hard to calculate $$\sum_{2n}{\binom{m}{n}}$$ (just because we know, how to denote the generating function, which keeps $a_{2n}$ stable and resets $a_{2n+1}$ to $0$,if we know the $\sum_{n \ge 0}{a_{n}z^{n}}$), but i got into troubles while trying to cope with some more complicated problems, such as presented above. 
Any help would be appreciated.

Consider that, by the discrete Fourier transform:
$$ 4\cdot\mathbb{1}_{n\equiv 0\pmod{4}} = 1^n+(-1)^n+i^n+(-i)^n \tag{1}$$
hence:
$$ \mathbb{1}_{n\equiv 2\pmod{4}} = \frac{1}{4}\left(1^n + (-1)^n - (-i)^n - i^n\right)\tag{2} $$
and:
$$\begin{eqnarray*} \sum_{n\equiv 2\pmod{4}}\binom{m}{n} &=& \frac{1}{4}\sum_{n=0}^{m}\binom{m}{n}\left(1^n+(-1)^n-i^n-(-i)^n\right)\\ &=& \frac{1}{4}\left(2^m-(1+i)^m-(1-i)^m\right).\tag{3}\end{eqnarray*}$$
