Prove the binomial sum $\sum_{k=0}^n k\binom{2n}{k} = \frac{n4^n}2$. Prove $\displaystyle \sum_{k=0}^n k\binom{2n}{k} = \frac{n4^n}2$.
This is from section 0.241 of Gradshteyn and Ryzhik's table of integrals.
I have managed to find a proof for this using the identity $k\binom nk = n \binom{n-1}{k-1}$.
Hence \begin{align*}
& \binom{2n}1+2\binom{2n}2+3\binom{2n}3+\cdots + n\binom{2n}n\\
&= 2n\binom{2n-1}0 + 2n\binom{2n-1}1 + 2n\binom{2n-1}2 + \cdots + 2n\binom{2n-1}{n-1}\\
&= 2n\bigg[\binom{2n-1}0 + \binom{2n-1}1 + \binom{2n-1}2 + \cdots + \binom{2n-1}{n-1}\bigg]\\
&= 2n\left(\frac{2^{2n-1}}2\right)\\
&= n \cdot 2^{2n-1}
\end{align*}
I was wondering if there was another method to do this (such as using calculus as the coefficients suggest this may work). I was unable to find a calculus solution as I don't think I can use symmetry to remove the second half of the terms.
 A: Combinatorial proof here: from $2n$ children we select some (not more than half shall be selected) to be choirs, in addition we select one child from the choirs to be the leader.
Left Hand Side
We select $k$ children where $k$ can be anywhere from $1$ to $n$. Then we select one leader from these $k$ children. The number of possibilities is given below:
$$
\begin{align}
\sum_{k=1}^{n}\binom{2n}{k}\cdot\binom{k}{1}&=\sum_{k=1}^{n}k\cdot\binom{2n}{k}\\
\\
&=\sum_{k=0}^{n}k\cdot\binom{2n}{k}
\end{align}
$$
Right Hand Side
Choose the leader first then each of the remaining $2n-1$ children either join the choir or don’t join. The number of possibilities is given below:
$$
2n\cdot 2^{2n-1}
$$
In half of these possibilities there are less than or equal to $n$ children in the choir and in the other half there are more than $n$ children in the choir. Therefore the number of possibilities we want is given below:
$$
\frac{1}{2}\times 2n\cdot 2^{2n-1}=\frac{n\cdot4^{n}}{2}
$$
Conclusion
Since both left and right hand side counts the same objects they must be equal
Edit
In case the last paragraph for the Right Hand Side is not obvious:
$$
\begin{align}
2^{2n-1}&=\sum_{i=0}^{2n-1}\binom{2n-1}{i}\\
\\
&=2\sum_{i=0}^{n-1}\binom{2n-1}{i}
\end{align}
$$
A: Too complex solution based on algebra.
Consider
$$k  \binom{2 n}{k} x^{k-1}=\Bigg[ \binom{2 n}{k} x^k\Bigg]'$$
$$S_n=\sum_{k=0}^n \binom{2 n}{k} x^k=(x+1)^{2 n}-\binom{2 n}{n+1} x^{n+1} \, _2F_1(1,1-n;n+2;-x)$$ where appears the gaussian hypergeometric function.
Computing the derivative with respect to $x$, we have
$$S'_n=2 n (x+1)^{2 n-1}-\frac{(n-1) \binom{2 n}{n+1} x^{n+1} \, _2F_1(2,2-n;n+3;-x)}{n+2}-$$ $$(n+1) \binom{2
   n}{n+1} x^n \, _2F_1(1,1-n;n+2;-x)$$
Now, making $x=1$ and using the properties of the gaussian hypergeometric function
$$S'_n=\sum_{k=0}^n k  \binom{2 n}{k} =n\,4^n-n\,2^{2 n-1} =n\,2^{2 n-1} $$
