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I'm looking for a proof of the identity:

$$\sum\limits_{k=0}^n \binom{n}{k}\binom{2k}{k}\frac{(-1)^k}{2^{2k}}=2^{-2n}\binom{2n}{n}$$

which I first saw in Mathematica while trying to evaluate an integral.

I've attempted using the decomposition of $\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1} $ with induction, since the first term on the RHS simplifies nicely in the sum. However, I have trouble proceeding from there.

My question may also be related to the summation $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ discussed here. That question-asker used $(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n} $ and the Cauchy product to show $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$.

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  • $\begingroup$ If you write it as $\sum_{k=0}^n\binom{n}{k}\binom{2k}{k}2^{2n-2k}(-1)^k=\binom{2n}{n}$, left side looks like some form of combinatorial argument could be used (where the right side is for example number of balanced binary strings of length $2n$). Maybe someone can finish. $\endgroup$ – Sil May 27 '18 at 8:38
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$$\frac1{\sqrt{1+x}}=\sum_{k=0}^\infty\frac{(-1)^k}{4^k}\binom{2k}kx^k$$ and $$(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}.$$ Your sum is the coefficient of $x^n$ in the product of these series, that is the coefficient of $x^n$ in $(1+x)^{n-1/2}$. This is $$\frac1{n!}\left(\frac{2n-1}2\frac{2n-3}2\cdots\frac32\frac12\right) =\frac{(2n)!}{4^nn!^2}=\frac1{4^n}\binom{2n}n.$$

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Hint. For $|x|<1$, consider the product $$(1+x)^n\cdot\frac{1}{\sqrt{1+x}}=\sum_{k=0}^n \binom{n}{k}x^k\cdot\sum_{k=0}^\infty \binom{2k}{k}\frac{(-1)^kx^k}{2^{2k}}.$$ What is the coefficient of $x^n$? Note that you can find it in two ways!

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  • $\begingroup$ Thank you for the response. I chose Lord Shark's answer because his choice to write the binomial expansion in terms of x^(n-k) really made it click how the coefficient on $x^n$ reduced to my series. $\endgroup$ – user196574 May 27 '18 at 6:40
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    $\begingroup$ @user196574 It's fine. Note that by the Cauchy product the coefficient of $x^n$ is $\sum_k a_{n-k}b_k$ where $a_{n-k}=\binom{n}{n-k}=\binom{n}{k}$. $\endgroup$ – Robert Z May 27 '18 at 6:45

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