I have a polynomial $f(n)$ defined by $f(n) = n^2 * (n + 1)$ where $n \in W$ (whole number 0,1,2 ..)
I have to somehow manipulate its term to zero i.e.
$a * f(n) + b * f(n-1) + c * f(n-3) \cdots + i * f(n-k) = 0$
How to find $a, b, c \cdots i$ in above equation. $a,b,c \cdots i \in I$
Note : This relation is a part of other recurrence relation. So I am trying to remove the $n$ terms to make it the recurrence free of $n$
We have the backward difference operator $$ \nabla\colon f \mapsto (n\mapsto f(n)-f(n-1)) $$ maps a polynomial of degree $d$ to a polynomial of degree $\leq d-1$. Since your starting $f(n)=n^2(n-1)$ is a polynomial of degree $3$, we have $\nabla^4 f=0$, i.e., $$\label{eq1} f(n)-4f(n-1)+6f(n-2)-4f(n-3)+f(n-4)=0\text{ for all }n\tag{1} $$ But the coefficient of $f(n-4)$ is $+1$, not $+4$. We can fix this either by multiplying \eqref{eq1} by 4 $$ 4f(n)-16f(n-1)+24f(n-2)-16f(n-3)+4f(n-4)=0 $$ or by adding 5 times $$\label{eq2} f(n-1)-4f(n-2)+6f(n-3)-4f(n-4)+f(n-5)=0\tag{2} $$ to equation \eqref{eq1}, yielding $$ f(n)+f(n-1)-14f(n-2)+26f(n-3)-19f(n-4)+5f(n-5) =0. $$
