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

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!
