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?
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1This can be rewritten as a sum over $n$ closed expressions which evaluate to algebraic numbers. This can be useful if $n$ is fixed or much smaller than $m$. However, this may have been your starting point from which you arrived at the sum in your question. For fixed $n$ with an explicit value, the sum could be made more explicit as well. From your question I cannot infer whether you are interested in such cases. – ccorn Jan 10 '14 at 14:29
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Relevant: This question. – ccorn Jan 10 '14 at 23:37
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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 $$ \sum_{s=0}^{2m} \binom{2m}{s} = 2^{2m}$$
ShreevatsaR
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@ccorn: Thanks, indeed it was very wrong; I have rewritten the answer completely. – ShreevatsaR Jan 10 '14 at 09:18
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@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 $$\begin{align}\sum_{k=-\lfloor m/n\rfloor}^{\lfloor m/n\rfloor}\binom{2m}{m+kn}&=\binom{2m}{m-n\lfloor m/n\rfloor}+\cdots+\binom{2m}{m+n\lfloor m/n\rfloor}\\&=\sum_{k=0}^{2m}\binom{2m}{k}-\left(\sum_{k=0}^{m-n\lfloor m/n\rfloor-1}\binom{2m}{k}\right)-\left(\sum_{k=m+n\lfloor m/n\rfloor+1}^{2m}\binom{2m}{k}\right)\\&=(1+1)^{2m}-2\left(\sum_{k=0}^{m-n\lfloor m/n\rfloor-1}\binom{2m}{k}\right)\\&=2^{2m}-2\sum_{k=0}^{m-n\lfloor m/n\rfloor-1}\binom{2m}{k}.\end{align}$$
mathlove
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1Wrong: In the question, the summand's lower binomial argument is restricted to be congruent to $m\pmod n$. Try $n=2$, $m=3,4$ as counterexamples. – ccorn Jan 10 '14 at 08:23
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Yes I too made the same mistake (don't know how), but this is very wrong as @ccorn says. – ShreevatsaR Jan 10 '14 at 09:18
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