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I am trying to show that for any positive integer m, $$\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$$

Intuitively this seems to be true, $m = 0$ both sides evaluate to $1$, $m = 1$ both sides evaluate to $-2$.

I was looking at this identity for a potential connection, but it seems not applicable here $$\binom{2n}{n} = \binom{n}{0}^{2} + \binom{n}{1}^{2} + \binom{n}{2}^{2} + ... + \binom{n}{n}^{2}$$

Also I was trying to find connections between this and the binomial theorem, but no luck

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  • $\begingroup$ I don't think so, I see the similarity, but it doesn't help me with my question. Would appreciate an answer to my question $\endgroup$
    – user722457
    May 25, 2020 at 16:15
  • $\begingroup$ @DanielWang What is your question? $\endgroup$
    – Phicar
    May 25, 2020 at 16:20
  • $\begingroup$ Showing that $\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$ $\endgroup$
    – user722457
    May 25, 2020 at 16:25
  • $\begingroup$ It's exactly the same question, isn't it? What is the difference? BTW, the last answer to the linked question is the one to look at -- very simple and elegant. $\endgroup$
    – saulspatz
    May 25, 2020 at 16:25
  • $\begingroup$ The RHS is the same, but the LHS isn't $\endgroup$
    – user722457
    May 25, 2020 at 16:26

1 Answer 1

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$(1+x)^{2m} \ =\ {2m \choose 0} + {2m \choose 1}x\ ..........\ {2m \choose 2m}x^{2m}$

$(x-1)^{2m} \ =\ {2m \choose 0}x^{2m}-{2m \choose 1}x^{2m-1}\ .......\ {2m \choose 2m}$

Observe coefficient of $x^{2m}$ in $(1+x)^{2m}(1-x)^{2m}$ is nothing but $\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2}$ = $ S $

$S$ = coefficent of $x^{2m}$ in $(1-x^2)^{2m}$ which easily can be seen by binomial theorm = $(-1)^m\binom{2m}{m}$

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