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Please feel free to close this is necessary as I didn't see exactly this question (some variations that I tried but didn't seem to apply. Prove: $$\sum_{k=0}^{n}{\binom{n}{k}^2}=\binom{2n}{n}$$ I figured trying to muscle through it by expanding and then multiplying each term by variants of 1 such as $\frac{n^2}{n^2}$ and $\frac{n^2(n-1)^2}{n^2(n-1)^2}$ to get common denominators, but that was NOT the way it seemed as it got ugly pretty quickly. Then I tried thinking up formulas of the binomial theorem that I could differentiate like other sums of binomial coefficient problems and i couldn't think of one. Any help here would be great.


marked as duplicate by Eleven-Eleven, Amzoti, 23rd, Micah, Jim Apr 26 '13 at 4:21

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  • $\begingroup$ Have you tried induction? $\endgroup$ – Ben West Apr 26 '13 at 3:03
  • $\begingroup$ hmmmm...I have not... $\endgroup$ – Eleven-Eleven Apr 26 '13 at 3:06
  • $\begingroup$ this has been asked. see here. $\endgroup$ – Coffee_Table Apr 26 '13 at 3:09
  • $\begingroup$ Thank you! I couldn't find it before i typed it. $\endgroup$ – Eleven-Eleven Apr 26 '13 at 3:20

This is an immediate consequence of Vandermonde's identity. For $m,n,r\in \mathbb{N}_0$,

$$ {m+n \choose r} = \sum_{k=0}^r {m\choose k}{n\choose r-k}$$

Now set $m=r=n$, and replace ${n\choose n-k}$ with ${n\choose k}$.



$$\left(\sum_{0\le r\le n}\binom nr x^r \right)\left(\sum_{0\le r\le n}\binom nr x^{n-r}\right)=\sum_{0\le r\le 2n}\binom {2n}rx^r$$

Compare the coefficients of $x^n$

  • $\begingroup$ That's funny as I had a fading thought of trying something like that (not the second part, but the first part, since i know of other results using the binomial theorem). Thanks @lab! $\endgroup$ – Eleven-Eleven Apr 26 '13 at 3:26

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