Proving a summation involving binomial coefficients. I need to prove the following inductively: (http://upload.wikimedia.org/math/9/e/5/9e57871ba17c1ad48e01beb7e1bb3bb9.png)
$$\sum_{i=1}^{n} i{n \choose i} = n2^{n-1}$$
And for the life of me I can't figure out how to express the formula with n+1 in terms of n.
 A: Let $\displaystyle S=\sum_{i=0}^n i\binom ni,$
$\displaystyle  j=n-i\implies  S=\sum_{j=0}^n (n-j)\binom n{n-j}=\sum_{j=0}^n (n-j)\binom nj$  as $\displaystyle \binom nj=\binom n{n-j}$
$\displaystyle\implies S=\sum_{i=0}^n (n-i)\binom ni $
$\displaystyle\implies S+S=n\sum_{i=0}^n \binom ni$
Now, we can resort to this inductive proof and set $x=y=1$

Without Induction,
$$r\cdot\binom nr=r\cdot \frac{n\cdot(n-1)!}{r\cdot(r-1)!\cdot[(n-1)-(r-1)]!}=n\binom{n-1}{r-1}$$ for $0\le r\le n$ and $\binom nr=0$ for $r<0$ or $r>n$
Now, put  $a=b=1 $ in $\displaystyle (a+b)^{n-1}=\sum_{r=0}^{n-1}\binom{n-1}ra^{n-1}b^r$
A: ** I was going to withdraw this answer as it does not use induction, but decided to leave it anyway and take the criticism. Thanks lab bhattacharjee for setting me right** 
Start with
$$
(x+1)^n = \sum _i {n \choose i} x^i
$$
and differentiate with respect to $x$
$$
n(x+1)^{n-1} = \sum _i  {n \choose i} i  x^{i-1}
$$
Now set $x=1$ to get your identity
$$
n ~2^{n-1} = \sum _i  i~ {n \choose i} 
$$
