Discrete Maths Combinatorics $\sum_{k=1}^{n-1}[(-1)^k(n-k)k{n\choose_k}] =0$ I want to prove that:
$\sum_{k=1}^{n-1}[(-1)^k(n-k)k{n\choose_k}] =0$ 
I tried proving it using induction, but it didn't really work for me.
Could you help me with that?
Thanks.
 A: It suffices to see that:
$$(n-k)k{n\choose_k} = n(n-1) {n-2\choose k-1}\tag 1$$
and your sum could be transformed:
$$S=\sum_{k=1}^{n-1}[(-1)^k(n-k)k{n\choose_k}] =-n(n-1)\sum_{k=0}^{n-2}[(-1)^k{n-2\choose_k}]=-n(n-1)(1-1)^{n-2}  \tag 2$$
Edit
We can prove this kind of inequalities by differentiating the binomial identity, let:
$$f(x)=(x+1)^n=\sum_{k=0}^{n} x^k{n \choose k} \tag 3$$
and by computing the first derivative of $f$ we have :
$$ xf'(x)= \sum_{k=0}^{n} kx^k{n \choose k}\tag 4$$
Now if $x\neq 0$ then:
$$g(x)=\frac{xf'(x)}{x^n}= \sum_{k=0}^{n} kx^{k-n}{n \choose k}\tag 5$$
By derivation again we can find that:
$$x^{n+1}g'(x)= \sum_{k=0}^{n} k(k-n)x^{k}{n \choose k} \tag 6$$
hence one could see that: $$S=-(-1)^{n+1}g'(-1) \tag 7$$ and we can compute $g'(x)$ easily :
$$g'(x)=\frac{d}{dx}\frac{f'(x)}{x^{n-1}}=\frac{d}{dx}\left(n\left(1+\frac{1}{x}\right)^{n-1}\right) =n(n-1)\frac{-1}{x^2}\left(1+\frac{1}{x}\right)^{n-2}\tag 8$$
finally :
$$\boxed{\sum_{k=0}^{n} k(n-k)x^{k}{n \choose k}= n(n-1)x(x+1)^{n-2}} \tag 8$$
and take $x=-1$ in this last formula.
A: Hint: $k(n-k)\binom{n}{k}=n(n-1)\binom{n-2}{k-1}$.  You also need to assume that $n>2$.
