Whenever I solve a problem, I try if I can proceed on with the generalized statement (or expression) for the problem.

I came up with two interesting generalizations today.

  • The straight forward one - The number of ordered n-tuples of integers {xi}i=1n \{ x_i \}_{i=1}^n such that i=1nxii=1nxi=n1\sum_{i=1}^n x_i - \prod_{i=1}^n x_i = n-1 is equal to nkn+1nk-n+1 provided that 1xik1 \leq x_i \leq k

  • This one generalizes the type of problems where you need to find sum of binomial coefficients which are at certain gaps. By gaps I mean, suppose you need to find k=07(304k)\sum_{k=0}^7 {30 \choose 4k} as you can find here that binomial coefficients are at certain gaps of 44. (I hope I am able to explain my point clearly). So here's the generalization - for m,nNm,n \in \mathbb{N} such that mnm \leq n we have

k=0mkn(nmk)=2nmk=0m1cosn(πkm)einkmπ\sum_{k=0}^{mk \leq n} {n \choose mk} = \dfrac{2^n}{m} \sum_{k=0}^{m-1} \cos^{n} \left(\frac{\pi k}{m}\right) e^{i\frac{nk}{m} \pi}

Exercise -

  • Evaluate k=110(303k)\sum_{k=1}^{10} {30 \choose 3k}

  • Prove k=0n1(1)kcosn(πkn)=n2n1\sum_{k=0}^{n-1} (-1)^k \cos^n\left(\frac{\pi k}{n}\right) = \frac{n}{2^{n-1}}

  • Prove both the generalizations.

Note by Kishlaya Jaiswal
4 years ago

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For "sum of binomial coefficients which are at certain gaps", I think that the best approach is using Roots of Unity instead of trying to prove it by induction.

Calvin Lin Staff - 4 years ago

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Yes Sir, indeed that's the method even I used to prove it and henceforth tagged this post with RootsOfUnity filter ;)

Kishlaya Jaiswal - 4 years ago

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Just say a few words on how do we use roots of unity..

Vishal Yadav - 2 years, 4 months ago

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