# 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.

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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 $$\sum_{i=0}^n {n\choose i}i = n\sum_{i=0}^n {n-1 \choose i-1} = n\sum_{i=0}^{n-1} {n-1\choose i} = n2^{n-1}$$


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

• Nice answer! Much simpler than mine. – jiyanez Sep 14 '16 at 4:26
• @jiyanez Thanks. 'Symmetries' are always useful. – Felix Marin Sep 14 '16 at 4:30
• Could you please explain how n/i = n/n-i? – user193203821309 Sep 14 '16 at 12:03
• @differentialequation $${n \choose i} = {n! \over i!\,\left(n - i\right)!} = {n! \over \left(n - i\right)!\,\left[n - \left(n - i\right)\right]!} = {n \choose n - i}$$ – Felix Marin Sep 14 '16 at 19:46