Inductive proof that ${2n\choose n}=\sum{n\choose i}^2.$

I would like to prove inductively that $${2n\choose n}=\sum_{i=0}^n{n\choose i}^2.$$

I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively things like $${2n+2\choose n+1}={2n+2\over n+1}{2n+1\choose n}=2\cdot {2n+1\over n+1}{2n\choose n},$$

But I don't think it can lead me anywhere. I would like the proof to be as simple as possible.

• This is a special case of Chu-Vandermonde identity. Sometimes a proof by induction might be easier if you prove a more general result. Although I am not sure whether it helps in this case. Mar 4 '13 at 11:44
• I'll add a link to another question, where a combinatorial proof is given. Mar 4 '13 at 11:48
• This result is formulated too narrowly to have much chance of a inductive proof: knowing something about just the central binomial coefficients is insufficient in the induction because you don't have a useful recurrence realtion for just the central binomial coefficients. However, proving as Martin Sleziak suggested Chu-Vandermonde by induction (using Pascal's recurence) is a piece of cake. Mar 5 '13 at 11:00

Split the $$2n$$ elements into two groups of size $$n$$ Then the no. of ways of choosing $$n$$ from the $$2n$$ is the no. of ways of choosing $$i$$ from the 1st and $$n-i$$ from the 2nd and letting $$i$$ vary.