# Generalizations!

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 $$1 \leq x_i \leq k$$

 

• 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

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

  

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.

Note by Kishlaya Jaiswal
2 years, 11 months 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.

Staff - 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 1 year, 2 months ago