How to show $\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$ [duplicate]

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

• 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
– user722457
May 25, 2020 at 16:15
• @DanielWang What is your question? May 25, 2020 at 16:20
• Showing that $\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$
– user722457
May 25, 2020 at 16:25
• 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. May 25, 2020 at 16:25
• The RHS is the same, but the LHS isn't
– user722457
May 25, 2020 at 16:26

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