How do I find the exponential generating function of $a_n = (-1)^n \sum_{k=0}^{n}k \binom{n}{k}$?

Question

I'm learning Combinatorics and came across this question:

Find the exponential generating function of the following series $$a_n = (-1)^n \sum_{k=0}^{n}k \binom{n}{k}$$

I have problem with the part of the binom, because I know that when multiplying generating funciton the $n$ "part" is becoming $k \text{ or } n-k$ in the new sum from k to n (I hope you undertand what I mean). Anyway I would like to get some help

Answer 1

Recall that $$(1+x)^n=\sum_{k=0}^{n}{n\choose k}x^k.$$ Therefore $$a_n=(-1)^n\frac{d}{dx}(x+1)^n\bigg|_{x=1}=(-1)^nn(1+1)^{n-1}=(-1)^n2^{n-1}n.$$ Thus $$\text{EGF}_{(a_n)}(x)=\frac12\sum_{n\ge0}\frac{n}{n!}(-2x)^n=\frac12\sum_{n\ge1}\frac{(-2x)^n}{(n-1)!}=-x\sum_{n\ge0}\frac{(-2x)^n}{n!}=-xe^{-2x}.$$

Answer 2

Here is a brutal-force solution:

\begin{align*} \sum_{n=0}^{\infty} \frac{a_n}{n!}x^n &= \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{k}{k!(n-k)!} \right) (-x)^n \\ &= \sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{k}{k!}(-x)^k \cdot \frac{1}{(n-k)!} (-x)^{n-k} \\ &= \left( \sum_{k=0}^{\infty} \frac{k}{k!}(-x)^k \right) \left( \sum_{l=0}^{\infty} \frac{1}{l!}(-x)^l \right) \\ &= (-x e^{-x})(e^{-x}) \\ &= -x e^{-2x}. \end{align*}