$$ D_n = \left| \begin{matrix} n & -1 & -3 & 0 & 0 & \cdots & 0 & 0 & 0 \\ n & 1 & 2 & -3 & 0 & \cdots & 0 & 0 & 0 \\ n & 0 & 1 & 2 & -3 & \cdots & 0 & 0 & 0 \\ n & 0 & 0 & 1 & 2 & \cdots & 0 & 0 & 0 \\ n & 0 & 0 & 0 & 1 & \cdots & 0 & 0 & 0 \\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots \\ n & 0 & 0 & 0 & 0 & \cdots & 1 & 2 & -3 \\ n & 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 2 \\ n & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\ \end{matrix} \right| $$
Can anyone help me with this determinant? It is recursive formula but I can't get it.
Add the second up to the last rows to the first. Then the first row becomes $(n^2,0,\ldots,0)$. The new matrix is lower block-triangular and is in the form of $\pmatrix{n^2&0\\ n\mathbf1&A}$, where $A$ is by itself an upper triangular matrix with determinant 1. Hence $\det D_n=n^2$.