 Solution

Expand the tridiagonal determinant along its first row. The diagonal entry contributes $d{D}_{N-1}$. In the off-diagonal cofactor, expansion along its first column supplies the other factor $-k$, and the cofactor sign makes the net contribution $-k^2{D}_{N-2}$. ThereforeThe initial values are $D_1=d$ and $D_2=d^2-k^2$. Writing the same formula at $N=2$ forces $\boxed{D_0=1}$, the empty determinant convention. This is the tridiagonal determinant recurrence.