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Expanding $\det A^{(n)}$ along its last row gives the last diagonal contribution $a_n{d}_{n-1}$. The only other nonzero entry is $c_{n-1}$; expanding its minor along the last column contributesThus the tridiagonal determinant recurrence isso

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.

Let $f_j$ be the determinant of the leading $j\times j$ principal submatrix. Expansion along the last row gives an $a_j{f}_{j-1}$ term and a $c_{j-1}$ term with negative cofactor sign. In the latter minor, the final column has only $b_{j-1}$ nonzero, and deleting that row and column leaves the leading $(j-2)\times (j-2)$ principal submatrix. Thus the tridiagonal determinant recurrence isHere $f_0=1$ is the empty determinant convention, needed to include $f_2=a_1{a}_2-b_1{c}_1$. The derivation uses cofactor expansion, so the recurrence relation remains valid when coefficients or intermediate determinants vanish; it never divides by a pivot element.