Find the generating function of the sequence $a_n = \sum\limits_{k=0}^n k(k-1)$ 
Find the generating function of the sequence $ a_n =\sum\limits_{k=0}^n k(k-1)$

My try:
Let's assume $k(k-1)$ is genereated by $F(x)$ then $a_n$ is generated by $\frac{F(x)}{1-x}$ (that's a common trick).
So have we reduced the problem to: Find the generating function of $a_k = k(k-1)$.
Now, there's another trick: If $G(x)$ generates $b_k$ then $x\cdot G'(x)$ generates $c_k = n\cdot b_k$
If so, we need to find the generating function of $c_k = k-1$.
I'm not sure I'm on the right path, can you help me with it?
Thanks!

from your initial conditions, we have
$$a_n=\sum_{k=0}^nk^2-\sum_{k=0}^nk\\
=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\\
=\frac{n(n+1)(n-1)}{3}$$
thus the generating function is
$$G(a_n,x)=\frac{1}{3}\sum_{n=0}^\infty(n^3-n)x^n\\
=\frac{1}{3}\left[\sum_{n=0}^\infty n^3x^n-\sum_{n=0}^\infty nx^n\right]$$
other hand, we have
$$<1,1,1,1...>=\frac{1}{1-x}\\
\Rightarrow<1,2,3,4...>=\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}\\
\Rightarrow<0,1,2,3...>=x\cdot\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}$$
so $\sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}$.
Then
$$<1,4,9,16...>=\frac{d}{dx}\frac{x}{(1-x)^2}=\frac{1+x}{(1-x)^3}\\
\Rightarrow<0,1,4,9...>=x\cdot\frac{1+x}{(1-x)^3}=\frac{x(1+x)}{(1-x)^3}\\
\Rightarrow<1,2^3,3^3...>=\frac{d}{dx}\frac{x(1+x)}{(1-x)^3}=\frac{1+4x+x^2}{(1-x)^4}\\
\Rightarrow<0,1,2^3,3^3...>=x\cdot\frac{1+4x+x^2}{(1-x)^4}=\frac{x+4x^2+x^3}{(1-x)^4}$$
which means $\sum_{n=0}^\infty n^3x^n=\frac{x+4x^2+x^3}{(1-x)^4}$.
So the result is
$$G(a_n,x)=\frac{1}{3}\left[\frac{x+4x^2+x^3}{(1-x)^4}-\frac{x}{(1-x)^2}\right]\\
=\frac{1}{3}\frac{6x^2}{(1-x)^4}\\
=\frac{2x^2}{(1-x)^4}$$
