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I have to evaluate this expression: $\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}$,

(In the original question we had $\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{k}$)

this is what I have done:

$$\begin{aligned} \sum_n\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n & = \sum_k\sum_n\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n \\ & = \sum_k\binom{2k}{k}(-2)^{-k}\sum_n\binom{n}{k}x^n \\ & = \sum_k\binom{2k}{k}(-2)^{-k}\frac{x^k}{(1-x)^{k+1}} \\ & = \frac{1}{1-x}\sum_k\binom{2k}{k}(\frac{x}{2x-2})^k \end{aligned}$$

now we know that $\sum_k\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$ and so we get

$$ \sum_n\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n = \frac{1}{\sqrt{1-x^2}} $$

So the sum that I'm looking for is equal to $\sum_{k=0}^n\binom{-1/2}{k}(-1)^k\binom{-1/2}{n-k}$

there is a way to express this expression without using sums or menus signs?

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    There is a typo $(-2)^k \frac{x^k}{(1-x)^{k+1}} = \frac{(-2 x)^k}{(1-x)^{k+1}}$, not $\frac{1}{1-x} \left( \frac{-2x}{2x-2} \right)^k$.2011-11-02
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    @Sasha: I fixed it2011-11-02
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    Some people may not know what "Snake Oil" means. See section 4.3 of Wilf, generatingfunctionology: http://www.math.upenn.edu/~wilf/DownldGF.html2011-11-02
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    If this result is true: $$\sum_n\left(\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}\right)x^n = \frac{1}{\sqrt{1-x^2}}$$ Then you just want the coefficient of $x^n$ in $(1-x^2)^{-1/2}$. So don't we get: $$\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}=\begin{cases}0&n\text{ odd}\\ \binom{-1/2}{n/2}&n\text{ even} \end{cases}$$?2017-02-24

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