How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$? 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.

Use Gauss' trick for summing a sequence: Combine the sum term-by-term with its reversed-order self, noting that the binomial coefficients are symmetric.
$$S = \sum_{r=0}^n r { n \choose r } = \sum_{r=0}^n (n-r) {n \choose n-r }$$
So, 
$$\begin{eqnarray}
2 S &=& \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose n-r} \right)\\\\
 &=& \sum_{r=0}^n \left( r {n\choose r} + (n-r) {n \choose r} \right)\\\\
&=& \sum_{r=0}^n \left( n {n\choose r} \right)\\\\
&=& n \sum_{r=0}^n {n\choose r}\\\\
&=& n 2^n \hspace{0.1in} \text{, by familiar identity}
\end{eqnarray}$$

To use a specific example, say, $n= 4$. The "trick" is deal to add the sum to its reverse, so let's look at those sums ...
$$\begin{align}
\sum_{r=0}^4 r\binom{4}{r} &\quad=\quad 0 \binom{4}{0} + 1 \binom{4}{1} + 2 \binom{4}{2} + 3\binom{4}{3} + 4\binom{4}{4} \quad= S \\
\sum_{r=0}^4(4-r)\binom{4}{4-r} &\quad=\quad 4 \binom{4}{4} + 3 \binom{4}{3} + 2 \binom{4}{2} + 1 \binom{4}{1} + 0\binom{4}{0} \quad= S
\end{align}$$ 
The fact that the binomial coefficients are "symmetric" says that each such coefficient in the first sum matches the one below it in the second sum. Adding the sums "vertically" (that is, term-by-term), we have
$$\sum_{r=0}^4 \left( r + (4-r) \right) \binom{4}{r} \quad=\quad 4\binom{4}{0} + 4\binom{4}{1} + 4\binom{4}{2} + 4\binom{4}{3} + 4\binom{4}{4} \quad=\quad 2 S$$ 
The trick has given us a common multiplier ($4$, which is to say $n$) that we factor-out:
$$n \sum_{r=0}^4 \binom{4}{r} \quad=\quad 4\left(\;\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} \;\right) \quad=\quad 2 S$$ 
Finally, because we know the sum of binomial coefficients is an appropriate power of $2$ (why?), this gives
$$4\cdot 2^4 = 2 S \qquad\to\qquad S = 4\cdot 2^3 = n\cdot 2^{n-1}$$
