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

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.
