combinatorics - Changing indices of summation, closed form summation

Using the fact that $\sum_{k=0}^{n}k(k+1)=\frac{n(n+1)(n+2)}{3}$, give a closed form sum for:

$\sum_{k=2n+1}^{3n}k(k+1)$.

So I made the substitution $i=k-(2n+1)$, and the new summation I wrote is:

$\sum_{i=0}^{n-1}(i+2n+1)(i+2n+2)$.

How do I give a closed form sum for this?

My guess is $2n+1$ is just a constant so if we define $A:=2n+1$, we have $\sum_{i=0}^{n-1}(i+A)(i+A+1)$, which kind of looks of the form $\sum_{k=0}^{n}k(k+1)=\frac{n(n+1)(n+2)}{3}$. Where do I go from here? I tried expanding the polynomial and separated the summation across the monomials but I end up with summations like $\sum_{i=0}^{n-1}A^2$ which I don't know how to evaluate since the index of summation is not present.