Sum with non unit increment

Question

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.

Answer 1

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: $$\begin{eqnarray*} \sum_{n\equiv 2\pmod{4}}\binom{m}{n} &=& \frac{1}{4}\sum_{n=0}^{m}\binom{m}{n}\left(1^n+(-1)^n-i^n-(-i)^n\right)\\ &=& \frac{1}{4}\left(2^m-(1+i)^m-(1-i)^m\right).\tag{3}\end{eqnarray*}$$

Answer 2

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 $4q+2$.