I am to derive the formula for $15,66,189,420, 795,\dots$
I have the common pattern as $3m(2n+1)$ where $n$ and $m$ are integers. For example,
$66=3\cdot2\cdot(2\cdot5+1)$
Will this end up in me deriving the correct formula or is there another pattern? Also could someone help me with the next step.
You seem to be on the right track. The numbers are
$$3\cdot 1\cdot 5$$
$$3\cdot 2\cdot 11$$
$$3\cdot 3\cdot 21$$
$$3\cdot 4\cdot 35$$
$$3\cdot 5\cdot 53$$
The differences in the last column are $6,10,14,18$ , forming an arithmetic progression. So, we have $$a_n=3n\cdot (2n^2+3)=6n^3+9n$$