Prove the identity $ \sum_{i=0}^n {n \choose i} i = n2^{n-1} $ with the identity $\sum_{k=0}^n \binom{n}{k}=2^n.$ 
Prove the identity $ \sum_{i=0}^n {n \choose i} i = n2^{n-1} $ with the identity $\sum_{k=0}^n \binom{n}{k}=2^n.$

I have already used calculus (differentiating both sides of the original identity) as one method, but I need help trying to do another. 
 A: You can do this by simply manipulating the combinatorial term.
$${n \choose i} i = \frac{i\cdot n!}{(n-i)!\cdot i!} = \frac{n!}{(n-i)!\cdot (i-1)!} = n\frac{(n-1)!}{((n-1)-(i-1))!\cdot (i-1)!}$$
(Notice that $((n-1)-(i-1)) = (n-i)$).
Then $$\sum_{i=0}^n {n\choose i}i = n\sum_{i=0}^n {n-1 \choose i-1} = n\sum_{i=0}^{n-1} {n-1\choose i} = n2^{n-1}$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\sum_{i = 0}^{n}{n \choose i}i} & =
\half\bracks{\sum_{i = 0}^{n}{n \choose i}i +
\sum_{i = 0}^{n}{n \choose n - i}\pars{n - i}}\qquad\pars{~Reflection~}
\\[5mm] & =
\half\,n\sum_{i = 0}^{n}{n \choose i}\qquad\qquad\qquad\pars{~\mbox{because}\
{n \choose i} = {n \choose n - i}~}
\\[5mm] & =
\half\,n\pars{2^{n}} = \color{#f00}{n\,2^{n - 1}}
\end{align}

Note that
  $\ds{\sum_{i = 0}^{n}a_{i} = \sum_{i = -n}^{0}a_{i + n} =
\sum_{i = n}^{0}a_{-i + n} = \sum_{i = 0}^{n}a_{n - i}\implies
\sum_{i = 0}^{n}a_{i} = \half\sum_{i = 0}^{n}\pars{a_{i} + a_{n - i}}}$.

