Find the sum of following series $\sum_{k=0}^{n}k\binom{2n+1}{k}$.
I don't know about correct answer but here's what I did : $$ \sum_{k=0}^{n}k\binom{2n+1}{k}=\sum_{k=0}^{n}k\frac{2n+1}{k}\binom{2n}{k-1}=(2n+1)\sum_{k=0}^{n}\binom{2n}{k-1}. $$
Now,
$$4^{n}=\binom{2n}{0}+\binom{2n}{1}+...+\binom{2n}{n-1}+\binom{2n}{n}+\binom{2n}{n+1}+...+\binom{2n}{2n}=2(\binom{2n}{0}+...+\binom{2n}{n-1})+\binom{2n}{n} \Rightarrow \sum_{k=0}^{n}k\binom{2n+1}{k}=(2n+1)(\frac{4^{n}-\binom{2n}{n}}{2}) $$
Is my approach correct?
