Trying to find $\sum\limits_{k=0}^n k \binom{n}{k}$ 
Possible Duplicate:
How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$? 

$$\begin{align}
&\sum_{k=0}^n k \binom{n}{k} =\\ 
&\sum_{k=0}^n k \frac{n!}{k!(n-k)!} =\\
&\sum_{k=0}^n k \frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!} = \\
&n\sum_{k=0}^n \binom{n-1}{k-1} =\\
&n\sum_{k=0}^{n-1} \binom{n}{k} + n \binom{n-1}{-1} =\\
&n2^{n-1} + n \binom{n-1}{-1}
\end{align}$$


*

*Do I have any mistake?  

*How can I handle the last term?


(Presumptive) Source: Theoretical Exercise 1.12(a), P18, A First Course in Pr, 8th Ed, by S Ross
 A: By convention $\binom{n}k=0$ if $k$ is a negative integer, so your last line is simply $$n\sum_{k=0}^n\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}\binom{n}k=n2^{n-1}\;.$$ Everything else is fine.
By the way, there is also a combinatorial way to see that $k\binom{n}k=n\binom{n-1}{k-1}$: the lefthand side counts the ways to choose a $k$-person committee from a group of $n$ people and then choose one of the $k$ to be chairman; the righthand side counts the number of ways to select a chairman ($n$) and then the other $k-1$ members of the committee.
A: If you know some elementary calculus, there's a simple way to do this.
By the Binomial theorem we have
$$
(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^k
$$
Differentiate both sides with respect to $x$:
$$
n(x+1)^{n-1}=\sum_{k=0}^n\binom{n}{k}kx^{k-1}
$$
Set $x=1$
$$
n2^{n-1}=\sum_{k=0}^n k\binom{n}{k}
$$
A: The annoying $\binom{n-1}{-1}$  at the end can be most easily avoided by starting with the observation that 
$$\sum_0^nk\binom{n}{k}=\sum_1^nk\binom{n}{k}.$$
Remark: For another way of summing the series, let $X$ be the number of tosses when a fair coin is tossed $n$ times. By symmetry we have $E(X)=n/2$.  But 
$$E(X)=\sum_0^n k\binom{n}{k}\frac{1}{2^n}.$$
The result follows immediately. A mean proof!
A: As an completely different method to compute the sum (though that's not what was asked for):
Note that the derivative of $$\sum_{k=0}^n {n\choose k}x^k=(1+x)^n$$
is 
$$\sum_{k=0}^n k {n\choose k}x^{k-1}=n(1+x)^{n-1}$$
and we want the value at $x=1$.
