TITLE: How can I solve $\sum\limits_{i = 1}^k i \binom{k}{i-1}$
QUESTION [6 upvotes]: If anyone could help with the steps in solving this summation.  I've been playing with it most of the day.  It started from me trying to prove $\sum\limits_{i=0}^{k+1} \left(i\binom{k+1}{i}\right) = (k+1)2^k$.  I used Pascal's rule to simplify this a little bit, but I think I've just been looking at it too long, and thought I'd see if anyone had any good initial hints.  Thanks in advance for any help.

REPLY [5 votes]: The following is a mean proof.
If we toss a fair coin $n$ times, the mean number of heads, by symmetry, is $\dfrac{n}{2}$.
But the mean is also
$$\sum_{i=0}^n i\binom{n}{i}\frac{1}{2^n}.$$
Remark: Minor manipulation deals with the variants mentioned in the OP. For example, changing $i\binom{n}{i}$ to $(i+1)\binom{n}{i}$ adds $2^n$ to the sum.