TITLE: Why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0$?
QUESTION [12 upvotes]: I know that the expansion of $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k}$ equals to zero. But why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k}$ also equal to zero for $n \geq 2$?
I've been using the first to derive the second, but it ended with no clue at all. Anyone know about how to derive this formula?
$$\displaystyle \sum_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0 .$$

REPLY [7 votes]: Here's a purely combinatorial proof that doesn't reduce the sum to the known identity $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k} = 0$.
The quantity $\binom{n}{k}k$ counts the number of ways to partition people numbered $\{1, 2, \ldots, n\}$ into a chaired committee $A$ of size $k$ and an unchaired committee $B$ of size $n-k$.  Given a particular commmittee pair $(A,B)$, let $x$ be the highest-numbered person in either committee who is not the chair of $A$.  Move $x$ to the other committee.  This mapping is defined for all pairs of committees when $n >1$, is its own inverse (and so is one-to-one), and changes the parity on committee pairs.  Thus, for $n > 1$, there are as many committee pairs with even parity as there are with odd parity.  In other words, $$\sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} k = 0$$
when $n > 1$.