$\sum(-1)^{k}k^{2}C_{n}^{k}$ I'm asked to compute $\sum^{n}_{k=1}(-1)^{k}k^{2}C_{n}^{k}$ for $n\geqslant 1$. I tried to use generating functions to solve the problem: $A(x):=\sum^{n}_{k=1}k^{2}C_{n}^{k}x^{k}$. So, I need to compute $A(-1)$. 
For $n=1$ $A(-1)=-1$, for $n=2$ $A(-1)=2$, for $n=2$ $A(-1)=2$, for $n=3,4,5,6,7,...(?)$ $A(-1)=0$. It seems that the answer is $0$ for $n\geqslant 3$...
Thank you in advance!

Computing
$$\sum_{k=0}^n {n\choose k} (-1)^k k^2$$
we write
$$\sum_{k=0}^n {n\choose n-k} (-1)^k k^2
= [z^n] (1+z)^n \sum_{k=0}^n z^k (-1)^k k^2.$$
Here the coefficient extractor controls the range and we find
$$[z^n] (1+z)^n \sum_{k\ge 0} z^k (-1)^k k^2
= [z^n] (1+z)^n \frac{z(z-1)}{(1+z)^3}
\\ = [z^n] (1+z)^{n-3} (z^2-z).$$
This is zero when $n\ge 3$ because we are extracting the coefficient
$[z^n]$ from a polynomial in $z$ of degree $n-1.$ 
Continuing, it becomes
$$[z^n] z^2 (1+z)^{n-3} - [z^n] z (1+z)^{n-3}
= {n-3\choose n-2} - {n-3\choose n-1}.$$
This is again zero by inspection when $n\ge 3$ because
$(n-3)^\underline{n-2} = 0$ and $(n-3)^\underline{n-1} = 0.$ 
For $n=2$ we get
${-1\choose 0} - {-1\choose 1} = 1 - (-1) = 2$
and for $n=1$ we find
$- {-2\choose 0} = -1.$
