Find the explicit formula for the numbers $a_n$

Question

Find the explicit formula for the numbers $a_n$ if $a_0=0$ and $$a_{n+1}=(n+1)a_n+2(n+1)!. \quad n\gt 0$$

As it is suggested by @dxiv:
$\;\dfrac{a_{n+1}}{(n+1)!}=\dfrac{a_n}{n!}+2\,$

My attempt:
Let $$F(x)=\sum_{n\geq 0} \frac{a_nx^n}{n!}=a_1x+\frac{a_2x^2}{2!}+\cdots$$ Multiply both sides of the recurrence relation by $x^{n+1}$ and sum over all natural numbers $n$ to get
$$\sum_{n\geq 0}\frac{a_{n+1}x^{n+1}}{(n+1)!}=\sum_{n\geq 0}\frac{a_nx^{n+1}}{n!}+\sum_{n\geq 0}2x^{n+1} \tag{*}\\ \sum_{n\geq 0}\frac{a_{n+1}x^{n+1}}{(n+1)!}=a_1x+a_2x^2+\cdots=F(x) \\ \sum_{n\geq 0}\frac{a_nx^{n+1}}{n!}=xF(x) \\ \sum_{n\geq 0}2x^{n+1}=\frac {2x} {1-x}$$ Then $(*)$ becomes $$F(x)=xF(x)+\frac {2x} {1-x} \\ F(x)=\frac {2x}{(1-x)^2}$$

Answer 1

Hint:   write it as $\;\dfrac{a_{n+1}}{(n+1)!}=\dfrac{a_n}{n!}+2\,$, which says that $\,\dfrac{a_n}{n!}\,$ is an arithmetic progression.

Answer 2

The sequence grows like $n!$ and so the ordinary generating function $G(x)$ is not going to be helpful. Better to use the exponential generating function $F(x) := \sum_{n=0}^\infty a_n x^n/n!.$

Now $a_0/0!=0$ and $a_{n+1}/(n+1)!=a_n/n!+2$ which implies $a_n/n!=2n$ by induction and $$F(x)=\sum_{n=0}^\infty 2nx^n=2x/(1-x)^2.$$

The generating function (ordinary or exponential) did not help us in this particular example.