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.

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
$$S'_n=\sum_{k=0}^n k  \binom{2 n}{k} =n\,4^n-n\,2^{2 n-1} =n\,2^{2 n-1} $$
