<s>[INST] An interesting Sum involving Binomial Coefficients How would you evaluate
$$\sum _{ k=1 }^{ n } k\left( \begin{matrix} 2n \\ n+k \end{matrix} \right) $$
I tried using Vandermonde identity but I can't seem to nail it down.
 [/INST] The following proof uses complex  variable techniques and improves the
elementary one I posted earlier.   It serves to demonstrate the method
even though it requires somewhat more of an effort.

Suppose we seek to evaluate
$$\sum_{k=1}^n k {2n\choose n+k}.$$
Introduce
$${2n\choose n+k} = 
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n-k+1}}
\frac{1}{(1-z)^{n+k+1}} \; dz.$$
Observe that this is zero when $k\gt n$ so we may extend
$k$ to infinity to obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}}
\frac{1}{(1-z)^{n+1}}
\sum_{k\ge 1} k \frac{z^k}{(1-z)^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}}
\frac{1}{(1-z)^{n+1}}
\frac{z/(1-z)}{(1-z/(1-z))^2} 
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n}}
\frac{1}{(1-z)^{n}}
\frac{1}{(1-2z)^2} 
\; dz.$$
Now put $z(1-z)=w$ so that (observe that with $w=z+\cdots$ the image of $|z|=\epsilon$ with $\epsilon$ small is another closed circle-like contour which makes one turn and which we may certainly deform to obtain another circle $|w|=\gamma$)
$$z = \frac{1-\sqrt{1-4w}}{2}
\quad\text{and}\quad
(1-2z)^2 = 1-4w$$
and furthermore
$$dz = -\frac{1}{2} 
\times \frac{1}{2} \times (-4) \times (1-4w)^{-1/2} \; dw
=  (1-4w)^{-1/2} \; dw$$
to get for the integral
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{1}{w^n} \frac{1}{1-4w} 
(1-4w)^{-1/2} \; dw
= \frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{1}{w^n} \frac{1}{(1-4w)^{3/2}} \; dw.$$
This evaluates by inspection to
$$4^{n-1} {n-1+1/2\choose n-1}
= 4^{n-1} {n-1/2\choose n-1}
= \frac{4^{n-1}}{(n-1)!}
\prod_{q=0}^{n-2} (n-1/2-q)
\\ = \frac{2^{n-1}}{(n-1)!}
\prod_{q=0}^{n-2} (2n-2q-1)
= \frac{2^{n-1}}{(n-1)!}
\frac{(2n-1)!}{2^{n-1} (n-1)!}
\\ = \frac{n^2}{2n} {2n\choose n} 
= \frac{1}{2} n {2n\choose n}.$$
Here the mapping  from $z=0$ to $w=0$ determines the  choice of square
root. For the conditions on $\epsilon$ and $\gamma$ we have that for the series to converge we require $|z/(1-z)|\lt 1$ or $\epsilon/(1-\epsilon) \lt 1$ or $\epsilon \lt 1/2.$ The closest that the image contour of $|z|=\epsilon$ comes to the origin is $\epsilon-\epsilon^2$ so we choose $\gamma \lt \epsilon-\epsilon^2$ for example $\gamma = \epsilon^2-\epsilon^3.$ This also ensures that $\gamma \lt 1/4$ so $|w|=\gamma$ does not intersect the branch cut $[1/4,\infty)$ (and is contained in the image of $|z|=\epsilon$). For example $\epsilon = 1/3$ and  $\gamma = 2/27$ will work.
</s>