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The question is to prove:$$\sum\limits_{r=0}^n{n\choose r}^2={2n\choose n}$$ using the binomial expansion of $(1+x)^n$ and $(1+x^{-1})^n$.

So firstly: $$(1+x)^n=\sum_{i=0}^n\frac{n!.x^n}{i!.(n-i)!}$$ $$(1+x^{-1})^n=\sum_{i=0}^n\frac{n!.x^{-n}}{i!.(n-i)!}$$ and we can work out that: $$\sum_{r=0}^n\left[\begin{pmatrix}n\\r\end{pmatrix}\right]^2=\sum_{r=0}^n\left(\frac{n!}{r!(n-r)!}\right)^2$$ but I am unsure of where to go from here

Did
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Henry Lee
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  • https://math.stackexchange.com/questions/793256/non-inductive-not-combinatorial-proof-of-sum-i-mathop-0n-binom-n-i2 – lab bhattacharjee Dec 04 '18 at 14:40
  • thank you that is the answer i was looking for, although how does this require $(1+x^{-1})^n$ – Henry Lee Dec 04 '18 at 14:41
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    We have $(1+x)^n(1+x^{-1})^n = \frac{(1+x)^{2n}}{x^n}$. The constant term in the power-series of the left hand side is the sum (the sum of terms $x^k$ from the first factor times terms $x^{-k}$ from the second factor). The constant term in the power-series on the right hand side is the binomial coefficient. – Winther Dec 04 '18 at 14:54
  • No $x^{-1}$ is required. Simply note that $$(1+x)^{2n}=(1+x)^n(1+x)^n=\sum_k{n\choose k}x^k\sum_j{n\choose j}x^{n-j}=\sum_{k,j}{n\choose k}{n\choose j}x^{n+k-j}$$ hence the coefficient of the $x^n$ term in the product on the RHS is $$\sum_{k,j}{n\choose k}{n\choose j}\mathbf 1_{k=j}=\sum_k{n\choose k}^2$$ – Did Dec 04 '18 at 15:09

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