Proving that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$ I have to show that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$.
What I know is that $\sum_{k=0}^{k=n} \binom{n}{k} \cdot k = 2^{n -1} \cdot n$.
How do I proceed from there?

\begin{align}
\sum_{k=0}^n k\binom{2n}{k} 
&= \sum_{k=1}^n k\binom{2n}{k} \\
&= \sum_{k=1}^n 2n\binom{2n-1}{k-1} \\
&= n \sum_{k=1}^n \left(\binom{2n-1}{k-1} + \binom{2n-1}{k-1}\right) \\
&= n \sum_{k=1}^n \left(\binom{2n-1}{k-1} + \binom{2n-1}{2n-k}\right) \\
&= n \sum_{j=0}^{2n-1} \binom{2n-1}{j} \\
&= n \cdot 2^{2n-1}
\end{align}

Alternatively, a combinatorial proof is to count the number of committees of size at most $n$ with one chairperson from $2n$ people.  The LHS conditions on the size $k$ of the committee.  The RHS selects the chairperson (in $2n$ ways) and then any subset of size at most $n-1$ from the remaining $2n-1$ people (to see that there are $2^{2n-1}/2$ of these, consider complementary pairs).
