# Is $\sum_{k=0}^n \left(k+1\right)\left(C^n_k\right)^2 = \frac{n+2}{2} C^{2n}_n$ for any positive integer $n$? [duplicate]

This question already has an answer here:

Today I had a test about IMO which is pretty hard though, I have worked on a combinatoric question that I found a special formula: $$\sum_{k=0}^n \left(k+1\right)\left(C^n_k\right)^2 = \dfrac{n+2}{2} C^{2n}_n$$ I don't know it is true or not, but it seems true because I have tested for some small $$n$$ and it is true. I have thought of using binomial theorem and breaking the $$C^{n}_k$$, but I still can't prove it. Luckily, I can do the question without using this formula, but I still someone can prove this formula. Tips and comments are welcome too.

## marked as duplicate by Sil, Robert Z, hardmath, Feng Shao, nmasantaAug 25 at 3:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

Using $$[x^n]$$ as the coefficient-extractor operator (it returns the coefficient of $$x^n$$ in the Maclaurin series of its argument) we have $$\sum_{k=0}^{n}(k+1)\binom{n}{k}^2 =[x^n]\left[\left(\sum_{k=0}^{n}\binom{n}{k}(k+1)x^k\right)\cdot\left(\sum_{k=0}^{n}\binom{n}{k}x^k\right)\right]$$ hence the LHS can be written as $$[x^n]\left[(1+x)^n\cdot \frac{d}{dx}\sum_{k=0}^{n}\binom{n}{k}x^{k+1}\right]=[x^n]\left[(1+x)^n\cdot \frac{d}{dx}\left(x(1+x)^n\right)\right]$$ or as $$[x^n] \left[(1+x)^n\cdot\left((1+x)^n+nx(1+x)^{n-1}\right)\right]=[x^n](1+x)^{2n}+n[x^{n-1}](1+x)^{2n-1}$$ which, by the binomial theorem, equals $$\binom{2n}{n}+n\binom{2n-1}{n-1}=\left(1+\frac{n}{2}\right)\binom{2n}{n}$$ as is was to be shown.