I would like to compute: $$D_n=\begin{vmatrix}1&-2&&&&\\2&2&-2&&0&\\&6&3&-2&&\\&&\ddots&\ddots&\ddots&\\&0&&\ddots&\ddots&-2\\&&&&n(n-1)&n\end{vmatrix}$$
I don't see any recurrence... We can factor the $n$ in the last line, but after that, all what I get is messy.
Subtracting $n-1$-times the nth column form the $n-1$st columns yields $$ D_n = \det\pmatrix{* &* &* \\ * & 3(n-1) & -2\\ 0 & 0 & n } $$ entries with $*$ are unchanged. Then $$ D_n = n \det\pmatrix{* &* \\ * & 3(n-1)} = n \det\pmatrix{* & * & 0 \\* & * & -2 \\ *&* & 3(n-1)} \\ = n \left( \det\pmatrix{* & * & 0 \\* & * & -2 \\ *&* & n-1} + \det\pmatrix{* & * & 0 \\* & * & 0 \\ *&* & 2(n-1)}\right), $$ where in the last step I used that the determinant is linear with respect to the columns of the matrix. Now the first determinant in the equation is $D_{n-1}$, the second is $2(n-1)D_{n-1}$. This implies $$ D_n = n ( D_{n-1} + 2(n-1)D_{n-2}) $$ with $D_1=1$. Setting $D_0=1$ makes the formula also valid for $n=2$.