5

I've tried to look at it combinatorically but I couldn't find a good model for $\binom{2n}{k}^3$, I've also tried to solve it using induction but the changes in steps are too much to keep track of

Can any one give a hint on any kind of proof for this?

Arnaud Mortier
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A.A.
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  • How about start by expanding the meaning of the binomial coefficient? – Enrico M. Feb 20 '18 at 13:48
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    Also you can take a look here: https://math.stackexchange.com/questions/793256/non-inductive-not-combinatorial-proof-of-sum-i-mathop-0n-binom-n-i2?rq=1 – Enrico M. Feb 20 '18 at 13:49
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    Should $k$ be $i$? – Arnaud Mortier Feb 20 '18 at 13:49
  • @ArnaudMortier yes. Thanks I fixed it – A.A. Feb 20 '18 at 13:51
  • Related: https://math.stackexchange.com/questions/1084326/how-to-prove-sum-k-1nk-binomnk-n2n-1?rq=1 and https://math.stackexchange.com/questions/683733/sum-of-k-n-choose-k-is-n2n-1 – Mr Pie Feb 20 '18 at 13:52
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    Up to sign, the RHS is $3n\choose n$$2n\choose n$ – Arnaud Mortier Feb 20 '18 at 14:03
  • Also related https://math.stackexchange.com/questions/1652748/prove-binom3nn-n-n-frac3nnnn-is-always-divisible-by-6-when – Arnaud Mortier Feb 20 '18 at 14:06
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    This is called Dixon's identity. An illuminating, yet non-elementary way is to recognize the LHS as a $_3F_2$ (generalized) hypergeometric function. See https://en.wikipedia.org/wiki/Dixon%27s_identity. Alternatively, this seems provable using WZ-pairs. – pisco Feb 20 '18 at 14:59
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    COMMENT.- In order to apply induction it could be useful to consider $$\binom{2(n+1)}{i}=\binom{2n}{i-2}+2\binom{2n}{i-1}+\binom{2n}{i}$$ and $$\sum_{i=0}^{2n}{(-1)^i \binom{2n}{i}}=2\sum_{i=0}^{n-1}{(-1)^i \binom{2n}{i}}+\binom{2n}{n}$$ (no time now to try to solve the problem). – Piquito Feb 20 '18 at 14:56

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