Binomial sum to be reduced? Let $0<n<m$ be integers.
Could the sum $\sum_{k=-\lfloor m/n \rfloor}^{\lfloor m/n \rfloor}\binom{2m}{m+kn}$ be expressed in closed form, or at least somewhat simplified?
 A: No, there is no closed form and probably no simpler expression either.
Let's say $m = nq + r$, where $0 \le r < n$.
The smallest value that $m+kn$ takes is for $k = -\lfloor m/n \rfloor = -q$, at which it is $m - nq = r$.
The largest value it takes is for $k = \lfloor m/n \rfloor = q$, at which it is $m + nq = 2nq + r$. In between, it takes all values that are $r$ modulo $n$, namely $r + nk$, for $0 \le k \le 2q$. So your sum can also be written as
$$\sum_{k=0}^{2q} \binom{2m}{r + nk}$$
There is no simple expression for this, but for $m \gg n$, a crude approximation may be $2^{2m}/n$, as you're picking only one every $n$ terms of the sum
$$ \sum_{s=0}^{2m} \binom{2m}{s} = 2^{2m}$$
A: @Anonymous : This answer is not correct. I cannot delete this because you accepted this as an answer. 
You might not like this, but a bit simpler.
If $m/n$ is a natural number, then your sum is $2^{2m}$.
If $m/n$ is not a natural number, yours will be
$$\begin{align}\sum_{k=-\lfloor m/n\rfloor}^{\lfloor m/n\rfloor}\binom{2m}{m+kn}&=\binom{2m}{m-n\lfloor m/n\rfloor}+\cdots+\binom{2m}{m+n\lfloor m/n\rfloor}\\&=\sum_{k=0}^{2m}\binom{2m}{k}-\left(\sum_{k=0}^{m-n\lfloor m/n\rfloor-1}\binom{2m}{k}\right)-\left(\sum_{k=m+n\lfloor m/n\rfloor+1}^{2m}\binom{2m}{k}\right)\\&=(1+1)^{2m}-2\left(\sum_{k=0}^{m-n\lfloor m/n\rfloor-1}\binom{2m}{k}\right)\\&=2^{2m}-2\sum_{k=0}^{m-n\lfloor m/n\rfloor-1}\binom{2m}{k}.\end{align}$$
