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I would like some help to prove the following equality : $$\sum_{i=0}^n \binom{n}i^2=\binom{2n}n$$ I wanted to do a proof by induction : $$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=1}^{n+1} \binom{n+1}i^2=1 + \sum_{i=0}^{n} \binom{n+1}{i+1}^2=1+\sum_{i=0}^{n} \bigg(\binom{n+1}i+\binom{n}{i+1}\bigg)^2$$ $$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=0}^{n} \bigg(\binom{n}i^2+2\binom{n}i\binom{n}{i+1}+\binom{n}{i+1}^2\bigg)=1+\binom{2n}n+ \sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1} +(\sum_{i=0}^n \binom{n}i^2-1) $$ $$\sum_{i=0}^{n+1} \binom{n+1}i^2=2\binom{2n}n+\sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1}=2\bigg(\sum_{i=0}^n \binom{n}i^2(1+\frac{n-i}{i+1})\bigg)=2(n+1)\bigg(\sum_{i=0}^n \binom{n}i^2\frac{1}{i+1}\bigg)$$

But now I'm stuck.

Kilkik
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  • The $^2$ is inside of the sum so I don't think so – Kilkik Mar 01 '21 at 21:42
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    Does https://math.stackexchange.com/questions/320348/inductive-proof-that-2n-choose-n-sumn-choose-i2 answer your question? – vitamin d Mar 01 '21 at 21:44
  • vitamin d's link is informative. I'll also include the link within that link: https://math.stackexchange.com/questions/219928/inductive-proof-for-vandermondes-identity/219938#219938. What you have is a specific case of Vandermonde's identity with m=n=r. You can prove Vandermonde's identity with induction, but it would be difficult to prove your equality using induction. This is because if you prove a stronger result with induction, then your inductive hypothesis is stronger... – Kevin Long Mar 02 '21 at 00:19
  • ...You could finish your inductive proof with a combinatorial proof, but that combinatorial proof would be enough to prove the original claim without induction, so it kind of defeats the purpose. – Kevin Long Mar 02 '21 at 00:20

2 Answers2

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One way to proceed is to prove a more general identity by induction, and then deduce the identity in the problem statement as a corollary.

The more general identity is called "Vandermonde's Identity": $$\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} = \binom{m+n}{k} \tag{1}$$ for non-negative integers $k$, $m$ and $n$. We give a proof by induction over $n$. The case $n=0$ reduces to $\binom{m}{k} = \binom{m}{k}$. Now suppose $(1)$ holds for some $n \ge 0$. Then the $n+1$ case is $$\begin{align} \sum_{i=0}^k \binom{m}{i} \binom{n+1}{k-i} &= \sum_{i=0}^k \binom{m}{i} \left( \binom{n}{k-i-1} + \binom{n}{k-i} \right) \tag{2}\\ &=\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i-1} + \sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} \\ &=\binom{m+n}{k-1} + \binom{m+n}{k} \tag{3} \\ &= \binom{m+n+1}{k} \tag{4} \end{align}$$ This completes the proof by induction. At steps $(2)$ and $(4)$ we used the identity $$\binom{n}{m} = \binom{n-1}{m-1} + \binom{n-1}{m}$$ and at step $(3)$ we used the inductive hypothesis.

To prove $$\sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$$ from $(1)$, take the special case $k=n$, $m=n$, and apply the identity $$\binom{n}{i} = \binom{n}{n-i}$$

awkward
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  • @MarkusScheuer That was my initial thought too, but it now seems to me that at each step in the induction the relevant equations are true for all $m$ using well-known identities, so I don't see the need for induction on $m$. – awkward Mar 05 '21 at 13:58
  • @MarkusScheuer I'm still having difficulty seeing your point. Is your problem with the proof of Vandermonde's Identity, or is it with the final step to deduce the identity in the O.P.? If the latter, Vandermonde's is true for arbitrary $k,m,n$ so it is surely also true in the special case where $k=m=n$. – awkward Mar 05 '21 at 15:21
  • I thought about this approach while I was jogging today and now I see the point. Your answer is sound. Well done. (+1) – Markus Scheuer Mar 06 '21 at 16:34
  • @MarkusScheuer Thank you. I value your opinion. – awkward Mar 06 '21 at 17:37
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The binomial theorem says $$(1+x)^n=\sum^n_{i=0}\binom{n}{i}x^i,$$ and we know that $$(1+x)^n(1+x)^n=(1+x)^{2n}.$$ Comparing the coefficient of $x^n$, we get $$\sum^n_{i=0}\binom{n}{i}\binom{n}{n-i}=\sum^n_{i=0}\binom{n}{i}^2$$ on the LHS, and $\binom{2n}{n}$ on the RHS.

qfwfq
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