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 \( \{ x_i \}_{i=1}^n\) such that \[\sum_{i=1}^n x_i - \prod_{i=1}^n x_i = n-1\] is equal to \(nk-n+1\) provided that \(1 \leq x_i \leq k\)

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- 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 \(\sum_{k=0}^7 {30 \choose 4k}\) as you can find here that binomial coefficients are at certain gaps of \(4\). (I hope I am able to explain my point clearly). So here's the generalization - for \(m,n \in \mathbb{N}\) such that \(m \leq n\) we have

\[\sum_{k=0}^{mk \leq n} {n \choose mk} = \dfrac{2^n}{m} \sum_{k=0}^{m-1} \cos^{n} \left(\frac{\pi k}{m}\right) e^{i\frac{nk}{m} \pi}\]

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**Exercise -**

Evaluate \[\sum_{k=1}^{10} {30 \choose 3k}\]

Prove \[\sum_{k=0}^{n-1} (-1)^k \cos^n\left(\frac{\pi k}{n}\right) = \frac{n}{2^{n-1}}\]

Prove both the generalizations.

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TopNewestFor "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.

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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

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

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