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.

The following solution is supposed to be the simplest one.
Let's consider a complex number $a$ such that $a^{4}=1$.
So, denote $S_{n}=\sum_{s=0}{\binom{i}{m}}$. We know its generating function $A(s)=(1+s)^{m}$. So, the generating function for the sequence over all $i=4q, q\in \mathbb{Z}$ would look like this $B(s)=A(s)+A(as)+A(a^{2}s)+A(a^{3}s)=(1+s)^{n}+(1+as)^{n}+(1+a^{2}s)^{n}+(1+a^{3}s)^{n}$, $B(1)=2^{n}+(1+a)^{n}+(1+a^{2})^{n}+(1+a^{3})^{n}$. So, since we know how to evaluate the sum over $4q$, then we know how to deal with it over $4q+2$.
