We know that function: $$ f(n)=\sum_{k=0}^nk(k+1) $$ is a polynomial of third degree.
I am tasked to write this polynomial using interpolation.
How to do it? Should I just come up with 4 different nodes (to get polynomial of third degree) e.g. $1,\ 2,\ 3,\ 4$ and calculate their value with $f(n)$ and then just interpolate them? (I know Newton and Lagrange methods)
I'm not asking you to solve this, just tell whether my method is correct (or tell me another solution).
As this polynomial is unique, any polynomial interpolation method on at least four points will work. More than four is overkill (the higher degree coefficients will end-up being zero).
As by-hand Lagrange interpolation is not so fun (consider using the Neville scheme), you can take advantage of the fact that there is no constant term ($f(0)=0$), and interpolate
$$\frac{f(n)}n$$ on three points instead.
The method of indeterminate coefficients might be advantageous.
$$\begin{cases}a+b+c&=\dfrac21,\\4a+2b+c&=\dfrac82,\\9a+3b+c&=\dfrac{20}3.\end{cases}$$
Additional hint:
By integration, we see that the sum must be asymptotic to $\dfrac{n^3}3$, so that the leading coefficient is $\dfrac13$.
Then you can perform a two-points interpolation on
$$\frac{f(n)-\dfrac{n^3}3}{n}.$$
$$\begin{cases}a+b&=\dfrac{2-\dfrac13}1,\\2a+b&=\dfrac{8-\dfrac83}2.\end{cases}$$
The solution is quasi-immediate.
A better trick:
Notice that
$$k(k+1)(k+2)-(k-1)k(k+1)=3k(k+1).$$
Then by telescoping,
$$\sum_{k=1}^n k(k+1)=\dfrac{n(n+1)(n+2)-0\cdot1\cdot2}3.$$
The trick works with all raising factorials.