Let $p$ be natural number such that $p \geq 2$. And $q$ is natural number such that $ 0 \leq q \leq p-1 $.
What is $ \sum _{i \equiv q \pmod p} {n \choose i}$? I solved it for $p=2, 3, 4, 5, 7$, but I cannot generalize it. I solved it by using complex number $w$ such that $w^p=1$ and Binomial theorem. Is there better ways?
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1I think the method using the binomial theorem is the best, and it also works for $p$ which are not prime. – Angina Seng Jul 19 '20 at 06:05
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then is there no generalizing? – 추민서 Jul 19 '20 at 06:06
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@추민서 Do you mean generalizing past these $p$ using the binomial theorem, or generalizing past these $p$ using some other method? – Carl Schildkraut Jul 19 '20 at 06:09
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I want to generalizing with real numbers. – 추민서 Jul 19 '20 at 06:11
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1Complex numbers is indeed the way to go, and it generalizes to all cases. The fancy name for this technique is Characters (in representation theory). – Calvin Lin Jul 19 '20 at 06:23
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by using it, I can generalize it? When $p=7$, it is very complex to solve it. is there another solution using characters? – 추민서 Jul 19 '20 at 06:27
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The formulas necessarily become more complicated as $p$ increases. Basically it is a combination of the binomial theorem and a discrete Fourier transform. When looking at representations of a cyclic group (see Calvin Lin's comment) we can replace characters with a DFT. Actually that applies to all finite abelian groups, but that's not relevant here. – Jyrki Lahtonen Jul 19 '20 at 06:52
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I hate to blow my own trumpet, but the only example I can find quickly is an old answer of mine. The same technique applies to variants of the harmonic infinite series as long as the DFT-component causing divergence vanishes. – Jyrki Lahtonen Jul 19 '20 at 06:55
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This and this have examples with $p=3$. You know about those already. I am adding them just to link more related questions. – Jyrki Lahtonen Jul 19 '20 at 07:06