Evaluating an expression using snake oil 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?

Added: This post answers the original question, which is to compute $\sum_{k=0}^\infty \binom{n}{k} \binom{2k}{k} (-2)^k$.

Assuming the typo is corrected you correctly arrived at the result:
$$
   \sum_{n=0}^\infty c_n x^n = \frac{1}{\sqrt{(1-x)(1+7 x)}}
$$
Where $c_n = \sum_{k=0}^\infty \binom{n}{k} \binom{2k}{k} (-2)^k$. Using 
$$
  \frac{1}{\sqrt{1-x}} = \frac{1}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{\Gamma(n+1/2)}{n!} x^n
   \qquad
   \frac{1}{\sqrt{1+7x}} = \frac{1}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{\Gamma(n+1/2)}{n!} (-7 x)^n
$$
Thus
$$
  \frac{1}{\sqrt{(1-x)(1+7x)}} = \sum_{n=0}^\infty x^n \sum_{m=0}^n \frac{1}{\pi} (-7)^m \frac{\Gamma(m+1/2) \Gamma(n-m+1/2)}{m! (n-m)!} 
$$
The sum over $m$ is the closed form which can, alternatively, be represented as a hypergeometric polynomial, giving
$$ \begin{eqnarray}
  c_n &=& \sum_{m=0}^n \frac{1}{\pi} (-7)^m \frac{\Gamma(m+1/2) \Gamma(n-m+1/2)}{m! (n-m)!} \\ &=&
    \frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right) n!} {}_2 F_1\left( \frac{1}{2}, -n; \frac{1}{2}-n ; -7 \right) \\ &=&
 {}_2 F_1 \left( \frac{1}{2}, -n ; 1 ; 8 \right)
\end{eqnarray}
$$

Added: To address OP's skepticism that the above is the correct solution to the question posed, here is some numerical verification using Mathematica:
In[73]:= Table[{Sum[
   Binomial[n, k] Binomial[2 k, k] (-2)^k, {k, 0, n}], 
  Hypergeometric2F1[1/2, -n, 1, 8], (
  Gamma[1/2 (1 + 2 n)] Hypergeometric2F1[1/2, -n, 1/2 - n, -7])/(
  Sqrt[\[Pi]] Gamma[1 + n])}, {n, 1, 6}]

Out[73]= {{-3, -3, -3}, {17, 17, 17}, {-99, -99, -99}, {609, 609, 
  609}, {-3843, -3843, -3843}, {24689, 24689, 24689}}

