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I'd like to find a method to find the summation of binomial coefficients of the following form using complex numbers. I don't need a proof that it works, but I do want to know how to approach this problem rather than just the method itself.

Also, infinity in the summation just means 'as long as the binomial coefficient makes sense', i.e the bottom part doesn't exceed the top one.

Thanks!

$$\sum^\infty_{j = 1} {m \choose k + jl}$$ $m, l, k \in \Bbb N$.

source

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  • $\begingroup$ What is $l$ ? The writing is illegible. $\endgroup$ – A---B Nov 8 '17 at 18:45
  • $\begingroup$ A fixed natural number. $\endgroup$ – John Nov 8 '17 at 19:25
  • $\begingroup$ Related: math.stackexchange.com/q/918 $\endgroup$ – Grigory M Nov 8 '17 at 21:44
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You have $k+jl \le m$ so $j \le \dfrac{m-k}{l} $ or $j \le \lfloor \dfrac{m-k}{l} \rfloor $.

So your sum is actually $\sum^{\lfloor (m-k)/l \rfloor}_{j = 1} {m \choose k + jl} $.

If you want to do this analytically, you can consider this as a multisection of $\sum_{j=1}^m x^j {m \choose j} $ as $x \to 1$.

See here:

https://en.wikipedia.org/wiki/Series_multisection

or here

http://mathworld.wolfram.com/SeriesMultisection.html

In particular, this last reference states that

$\sum_{m=0}^{\infty} {n \choose t+sm} =\dfrac1{s}\sum_{j=0}^{s-1}2^n\cos^n\left(\dfrac{\pi j}{s} \right) \cos\left(\dfrac{\pi(n-2)tj}{s} \right) $ where integers $0 \le t \lt s$ and the sum can only be taken up to $t+sm \le n$.

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  • $\begingroup$ Sorry, I'd like a more elementary approach to it since I don't know what most of the things in the link mean, although I can sort of see similarities.. $\endgroup$ – John Nov 9 '17 at 1:11

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