For a particular problem, I reached until this point where eventually I have to prove this summation


$$
\frac{1}{n} \left ( \binom{2n}{n+1} + 2\binom{2n}{n+2} + 3\binom{2n}{n+3} + \dots + n\binom{2n}{2n} \right ) = \frac{1}{2}\binom{2n}{n}
$$


I've tried to form a differentiating function that would result in this but no luck with that.



Combinatorial Summation $\frac{1}{n} \sum_{k=n+1}^{2n} (k-n)\binom{2n}{k}$
The left hand side can be rewritten as
\begin{align\*}
\sum\_{j=0}^n\left(1-\frac{j}{n}\right)\binom{2n}{j}
&=\sum\_{j=0}^n\binom{2n}{j}-\sum\_{j=0}^n\frac{j}{n}\binom{2n}{j}\\[6pt]
&=\sum\_{j=0}^n\binom{2n}{j}-2\sum\_{j=1}^n\binom{2n-1}{j-1}\\[6pt]
&=\frac{2^{2n}+\binom{2n}{n}}{2}-2^{2n-1}\\[6pt]
&=\frac{1}{2}\binom{2n}{n}.
\end{align\*}
Done.


