Prove equality $\sum_{i=0}^{n}i{{n}\choose{i}}=2^{n-1}n$ Prove that:
$$\sum_{i=0}^{n}i{{n}\choose{i}}=2^{n-1}n$$
I tried to use binominal theorem, but I don't know how to get rid of the $i$ and substitution didn't help me either.
 A: Observe that, for $1\leq i \leq n$,
$$
i{{n}\choose{i}} = i \frac{n!}{i!(n-i)!} = n \cdot \frac{(n-1)!}{(i-1)! (n-i)!} = n {{n-1}\choose{i-1}},
$$
hence
$$
\sum_{i=0}^{n}i{{n}\choose{i}}
= \sum_{i=1}^{n}i{{n}\choose{i}}
= n \sum_{i=1}^{n} {{n-1}\choose{i-1}} = 
n \sum_{j=0}^{n-1} {{n-1}\choose{j}} =n\, 2^{n-1}.
$$
A: You can write $\binom{n}{i} = \frac{n}{i} \binom{n-1}{i-1}$ for $i > 0$. 
Also note that $\sum_{i=0}^{n} \binom{n}{i}$ is equal to $(1+x)^n$ when $x = 1$. 
So the sum can be written as:
$$n\sum_{i=1}^{n} \binom{n-1}{i-1} = n \,2^{n-1}$$
A: It's
$$n2^{n-1}=((1+x)^n)'_{x=1}=\left(\sum_{i=0}^nx^i\binom{n}{i}\right)'_{x=1}$$
A: By the binomial theorem, $(1+x)^n = \displaystyle \sum_{i=0}^{n}\binom{n}{i}x^i$ so after differentiating wrt $x$ throughout and setting $x=1$, the result follows.
A: $(1+x)^n = \sum_{i=0}^{n}\binom{n}{i}x^i.$
Differentiate with resp. to $x:$
$n(1+x)^{n-1} =\sum_{i=1}^{n}\binom{n}{i}ix^{i-1}.$
Set $x=1$:
$n2^{n-1} = \sum_{i=1}^{n}i\binom{n}{i}$.
A: Let me tell you about one better way to do this
$$(1+x)^n = {n \choose 0} + {n \choose 1} x + {n \choose 2} x^2 + \ldots$$
Just differentiate both sides with respect to $x$ and put $x =1$
