TITLE: How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$?
QUESTION [18 upvotes]: I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more general way.
The left side of an identity occurs while solving another problem (concerning binomial theorem) so I am more interested in deriving the right side from the left side, else I have to remember it now onward.
EDIT: I am more interested in an algebraic proof rather than combinatorial argument or something involving calculus (however I liked svenkatr and Bill Dubuque solution), hence I am removing the combinatorics tag.

REPLY [8 votes]: Here's a dirty trick.  $\frac{1}{2^n} \sum_{r \ge 0} r {n \choose r}$ is the expected size of a random subset of an $n$-element set.  But by linearity of expectation, this is $n$ times the probability that any given element is in a subset, which is $\frac{1}{2}$.  So
$$\frac{1}{2^n} \sum_{r \ge 0} r {n \choose r} = \frac{n}{2}.$$
Edit:  And as long as I have this written up somewhere, I might as well use it.  In this math.SE answer I prove the following.  If $a_n, b_n$ are sequences satisfying $b_n = \sum_{k=0}^n {n \choose k} a_k$, and if $A(x) = \sum_{n \ge 0} a_n x^n, B(x) = \sum_{n \ge 0} b_n x^n$, then
$$B(x) = \frac{1}{1 - x} A \left( \frac{x}{1 - x} \right).$$
In this example $a_n = n$.  One can prove by various arguments that in this case $A(x) = \frac{x}{(1 - x)^2}$; this is a special case of an identity I use in the above answer.  It follows that
$$B(x) = \frac{1}{1 - x} \left( \frac{ \frac{x}{1-x} }{ \left( 1 - \frac{x}{1-x} \right)^2 } \right) = \frac{x}{(1 - 2x)^2} = \frac{1}{2} \frac{2x}{(1 - 2x)^2} = \frac{1}{2} \sum_{n \ge 0} n \cdot 2^n x^n$$
as desired.