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.
 A: $$
\begin{align}
\sum_{k=1}^nk\binom{2n}{n+k}
&=\sum_{k=1}^n(n+k)\binom{2n}{n+k}-\sum_{k=1}^nn\binom{2n}{n+k}\tag{1}\\
&=\sum_{k=1}^n2n\binom{2n-1}{n+k-1}-\sum_{k=1}^nn\binom{2n}{n+k}\tag{2}\\
&=n2^{2n-1}-\frac n2\left(2^{2n}-\binom{2n}{n}\right)\tag{3}\\
&=\frac n2\binom{2n}{n}\tag{4}
\end{align}
$$
Explanation:
$(1)$: $k=(n+k)-n$
$(2)$: $k\binom{n}{k}=n\binom{n-1}{k-1}$
$(3)$: $\sum_{k=0}^n\binom{n}{k}=2^n$. The red parts below are in $(2)$
$\phantom{(3):}$$\underbrace{\binom{2n-1}{0}+\binom{2n-1}{1}+\cdots+\binom{2n-1}{n-1}}_{\text{half of $2^{2n-1}$}}+\underbrace{\color{#C00000}{\binom{2n-1}{n}+\binom{2n-1}{n+1}+\cdots+\binom{2n-1}{2n-1}}}_{\text{half of $2^{2n-1}$}}=2^{2n-1}$
$\phantom{(3):}$$\underbrace{\binom{2n}{0}+\binom{2n}{1}+\cdots+\binom{2n}{n-1}}_{\text{half of $2^{2n}-\binom{2n}{n}$}}+\binom{2n}{n}+\underbrace{\color{#C00000}{\binom{2n}{n+1}+\binom{2n}{n+2}+\cdots+\binom{2n}{2n}}}_{\text{half of $2^{2n}-\binom{2n}{n}$}}=2^{2n}$
$(4)$: cancel
A: A different computational approach:
$$\begin{align*}
\sum_{k=1}^nk\binom{2n}{n+k}&=\sum_{k=1}^nk\binom{2n}{n-k}\\
&=\sum_{k=0}^{n-1}(n-k)\binom{2n}k\\
&=n\sum_{k=0}^{n-1}\binom{2n}k-\sum_{k=0}^{n-1}k\binom{2n}k\\
&=n\sum_{k=0}^{n-1}\binom{2n}k-2n\sum_{k=0}^{n-1}\binom{2n-1}{k-1}\\
&=n\sum_{k=0}^{n-1}\binom{2n}k-2n\sum_{k=0}^{n-2}\binom{2n-1}k\\
&=n\sum_{k=0}^{n-1}\left(\binom{2n-1}{k-1}+\binom{2n-1}k\right)-2n\sum_{k=0}^{n-2}\binom{2n-1}k\\
&=n\sum_{k=0}^{n-2}\binom{2n-1}k+n\sum_{k=0}^{n-1}\binom{2n-1}k-2n\sum_{k=0}^{n-2}\binom{2n-1}k\\
&=n\sum_{k=0}^{n-1}\binom{2n-1}k-n\sum_{k=0}^{n-2}\binom{2n-1}k\\
&=n\binom{2n-1}{n-1}
\end{align*}$$
Note that
$$\binom{2n}n=\frac{2n}n\binom{2n-1}{n-1}\;,$$
so
$$n\binom{2n-1}{n-1}=\frac{n}2\binom{2n}n\;,$$
and this answers agrees with the others already posted.
A: Permit me to contribute an algebraic proof.
Suppose we seek to evaluate
$$\sum_{k=0}^n k {2n\choose n+k}
= \sum_{k=0}^n (n-k) {2n\choose 2n-k}
= \sum_{k=0}^n (n-k) {2n\choose k}
\\ = n \frac{1}{2} \left(2^{2n} - {2n\choose n}\right)
+ n {2n\choose n}
- \sum_{k=0}^n k {2n\choose k}
\\ = n \frac{1}{2} \left(2^{2n} + {2n\choose n}\right)
- \sum_{k=0}^n k {2n\choose k}
.$$
Now we have
$$\sum_{k=0}^n k {2n\choose k}
= \sum_{k=1}^n k {2n\choose k}
= 2n \sum_{k=1}^n {2n-1\choose k-1}
\\ = 2n \sum_{k=0}^{n-1} {2n-1\choose k}
= 2n \frac{1}{2} 2^{2n-1}.$$
Collecting everything we obtain
$$n \frac{1}{2} \left(2^{2n} + {2n\choose n}\right)
- n \frac{1}{2} 2^{2n}
= \frac{1}{2} n {2n\choose n}.$$
Remark. This would appear to be the technique used by
@robjohn.
A: 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.
A: Let
$$ S_n = \sum_{k=1}^{n} k\binom{2n}{n+k}=(2n)!\cdot\sum_{k=1}^{n}\frac{k}{(n+k)!(n-k)!}.\tag{1}$$
We have:
$$\begin{eqnarray*} 2S_n &=& (2n)!\cdot \sum_{k=1}^{n}\frac{(n+k)-(n-k)}{(n+k)!(n-k)!}\\ &=& (2n)!\cdot \sum_{k=1}^{n}\left(\frac{1}{(n+k-1)!(n-k)!}-\frac{1}{(n+k)!(n-k-1)!}\right)\\&=&(2n)\cdot \sum_{k=1}^{n}\left(\binom{2n-1}{n+k-1}-\binom{2n-1}{n+k}\right)\tag{2}\end{eqnarray*} $$
but the last sum is a telescopic sum, that leads to:

$$ 2S_n = \color{red}{n\binom{2n}{n}}.\tag{3} $$

