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?
 A: \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).
A: $$\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+...+\dbinom{2n}{2n-1}+\dbinom{2n}{2n}=2^{2n}$$
$$\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+...+\dbinom{2n}{n-1}+\dbinom{2n}{n}=2^{2n-1}$$
$$\dbinom{n}{k}=\dbinom{n}{n-k}$$
$$k\dbinom{n}{k}+(n-k)\dbinom{n}{n-k}=n\dbinom{n}{k}$$
$$0\dbinom{2n}{0}+1\dbinom{2n}{1}+2\dbinom{2n}{2}+...+(2n-1)\dbinom{2n}{2n-1}+2n\dbinom{2n}{2n}=2n\dbinom{2n}{0}+2n\dbinom{2n}{1}+2n\dbinom{2n}{2}+...+2n\dbinom{2n}{n-1}+2n\dbinom{2n}{n}=2n(\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+...+\dbinom{2n}{n-1}+\dbinom{2n}{n})=2n\times 2^{2n-1}$$
Hence
$$n\dbinom{2n}{0}+n\dbinom{2n}{1}+n\dbinom{2n}{2}+...+n\dbinom{2n}{n-1}+n\dbinom{2n}{n}=n\times2^{2n-1}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{\sum_{k = 0}^{n}{2n \choose k}k = 2^{2n -1}\,\,n}:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{n}{2n \choose k}k} =
\sum_{k = 1}^{n}
{\pars{2n}! \over  \pars{k - 1}!\pars{2n - k}!}
\\[5mm] = &\
\sum_{k = 0}^{n - 1}
{\pars{2n}! \over  k!\pars{2n - k - 1}!} =
\color{red}{2n\sum_{k = 0}^{n - 1}{2n - 1 \choose  k}}
\\[5mm] = &\
2n\
\underbrace{\sum_{k = 0}^{2n - 1}{2n - 1 \choose  k}}
_{\ds{2^{2n - 1}}}\ -\
2n\sum_{k = n}^{2n - 1}{2n - 1 \choose  k}
\\[5mm] = &\
2^{2n}\,n - 2n\sum_{k = 0}^{n - 1}{2n - 1 \choose  k + n}
=
2^{2n}\,n -
2n\sum_{k = 0}^{n - 1}{2n - 1 \choose  n - 1 - k}
\\[5mm] = &\
2^{2n}\,n -
\color{red}{2n\sum_{k = 0}^{n - 1}
{2n - 1 \choose  k}}
\end{align}

\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{n}{2n \choose k}k} = \color{red}{2n\sum_{k = 0}^{n - 1}
{2n - 1 \choose  k}} = 
{2^{2n}\,n \over 2} = \bbx{2^{2n - 1}\,n} \\ &
\end{align}
