how to find this generating function this is the power series:
$$\sum_{i=0}^\infty n(n-1)^2 (n-2) z^n.$$
how can I find a generating function from it? I could use the third derivative but the $n-1$ is squared so I don't know what to do..
 A: $$
\begin{eqnarray}
\sum_{n=0}^\infty n(n-1)^2 (n-2) z^n
&=&
z^3\sum_{n=0}^\infty n(n-1)^2 (n-2) z^{n-3}
\\
&=&
z^3\frac{\mathrm d^3}{\mathrm dz^3}\sum_{n=0}^\infty(n-1)z^n
\\
&=&
z^3\frac{\mathrm d^3}{\mathrm dz^3}z^2\sum_{n=0}^\infty(n-1)z^{n-2}
\\
&=&
z^3\frac{\mathrm d^3}{\mathrm dz^3}z^2\frac{\mathrm d}{\mathrm dz}\sum_{n=0}^\infty z^{n-1}
\\
&=&
z^3\frac{\mathrm d^3}{\mathrm dz^3}z^2\frac{\mathrm d}{\mathrm dz}\frac1{z(1-z)}
\\
&=&
\frac{12z^3(1+z)}{(1-z)^5}
\end{eqnarray}
$$
A: At wolframalpha.com enter 
Sum n (n - 1)^2 (n - 2) z^n, n, 0, Infinity 

The series sum appears in an instant.
A: We can use
$$
\sum_{n=0}^\infty\binom{n}{k}z^n=\frac{z^k}{(1-z)^{k+1}}\tag{1}
$$
which follows from differentiating $\sum\limits_{n=0}^\infty z^n=\frac{1}{1-z}$ repeatedly $k$ times and multiplying by $\dfrac{z^k}{k!}$, and
$$
n(n-1)^2 (n-2)=n(n-1)(n-2)(n-3)+2n(n-1)(n-2)\tag{2}
$$
which is an example of the fact that any polynomial in $n$ can be written as a sum of $\binom{n}{k}$, we get
$$
\begin{align}
f(z)
&=\sum_{n=0}^\infty n(n-1)^2 (n-2) z^n\\
&=\sum_{n=0}^\infty n(n-1)(n-2)(n-3)z^n + 2n(n-1)(n-2)z^n\\
&=\sum_{n=0}^\infty 24\binom{n}{4}z^n + 12\binom{n}{3}z^n\\
&=\frac{24z^4}{(1-z)^5}+\frac{12z^3}{(1-z)^4}\\
&=\frac{12z^3(1+z)}{(1-z)^5}\tag{3}
\end{align}
$$
A: $$n*(n-1)^2 *(n-2) = n*(n-1)*(n-2)*(n-3+2)$$
$$ = n*(n-1)*(n-2)*(n-3) + 2*n*(n-1)*(n-2)$$
Basically, the solution is z^3 multiplied by the fourth derivative plus two time the third derivative of $$\sum z^n = 1/(1-z)$$
A: $$f(z)=\sum n(n-1)^2(n-2)z^n=12z^3+72z^4+240z^5+600z^6+1260z^7+\dots$$
$$(1-z)f(z)=12z^3+60z^4+168z^5+360z^6+660z^7+\dots$$
$$(1-z)^2f(z)=12z^3+48z^4+108z^5+192z^6+300z^7+\dots$$
$$(1-z)^3f(z)=12z^3+36z^4+60z^5+84z^6+108z^7+\dots$$
$$(1-z)^4f(z)=12z^3+24z^4+24z^5+24z^6+24z^7+\dots$$ 
From there, it should be easy. 
EDIT:
Note that this method will work for $\sum h(n)z^n$ for any polynomial $h$. If $h$ has degree $d$, then $(1-z)^{d+1}f(z)$ will be a polynomial, so you get your formula for $f(z)$. Note also that it requires no calculus, and is completely mechanical, needing no clever observations on ways to rewrite $h$. 
A: If you expand the terms you have $\sum_{i=0}^\infty n*(n-1)^2*(n-2)*z^n=\sum_{i=0}^\infty (n^4-4n^3+4n^2-2n)z^n$  and you can deal with each term.  Alternately $\sum_{i=0}^\infty n*(n-1)^2*(n-2)*z^n=\sum_{i=0}^\infty z^3\frac{d^2}{dz^2}z\frac {d^2}{dz^2}z^n$
A: In general, if $A(z) = \sum_{n \ge 0} a_n z^n$, you get:
$$
z \frac{\mathrm{d}}{\mathrm{d} z} A(z) = \sum_{n \ge 0} n a_n z^n
$$
So, using $\mathrm{D}$ for "derivative," and $p(n)$ a polynomial:
$$
\sum_{n \ge 0} p(n) a_n z^n = p(z \mathrm{D}) A(z)
$$
This gives a general method to get sums as yours. Just take care, e.g. $(z \mathrm{D})^2 \ne z^2 \mathrm{D}^2$.
