How do you express $j^3$ $-2j^2$ $+j$ in falling factorials?

Question

How do you express $j^3$ $-2j^2$ $+j$ in falling factorials?

So that you can calculate easily $\sum_{j=0}^n$ $j^3$$-2j^2$$+j$ ?

Answer 1

$$\begin{align}j^3-2j^2+j&=j(j^2-2j+1)\\ &=j(j-1)^2\\ &=j(j-1)\left[(j-2)+1\right]\\ &= j(j-1)(j-2) + j(j-1)\end{align}$$

The brute force approach is to write:

$$j^3-2j^2+j = a_3j(j-1)(j-2) + a_2j(j-1)+a_1j + a_0$$

Note that $a_3$ must be $1$, so now you need:

$$j^3-2j^2+j-j(j-1)(j-2)=j^2-j=a_2j(j-1)+a_1j+a_0$$

Now you see that $a_2=1, a_1=0, a_0=0$. (If it wasn't obvious at this point, you could still immediately see what $a_2$ was, then subtract again...)