= Solution
Substitute $D_N=\beta^N$ into the recurrence. Since $k>0$, its roots are nonzero and satisfy
$$
\beta^2-d\beta+k^2=0,\qquad\boxed{\beta_1+\beta_2=d,\quad\beta_1\beta_2=k^2.}
$$
For distinct roots, the general recurrence solution is $D_N=c_1\beta_1^N+c_2\beta_2^N$. The initial values require $c_1+c_2=1$ and $c_1\beta_1+c_2\beta_2=d=\beta_1+\beta_2$, giving
$$
\boxed{c_1=\frac{\beta_1}{\beta_1-\beta_2},\qquad c_2=-\frac{\beta_2}{\beta_1-\beta_2},\qquad D_N=\frac{\beta_1^{N+1}-\beta_2^{N+1}}{\beta_1-\beta_2}.}
$$
At a repeated root, $d=\pm2k$ and $\beta_1=\beta_2=\beta=\pm k$, this quotient must be replaced by its limit:
$$
\boxed{D_N=(N+1)\beta^N.}
$$
It also follows by solving the repeated-root recurrence as $(c_1+c_2N)\beta^N$. This exceptional case is necessary to avoid introducing a false root in the next part; it is included in the <fixed-end spring-chain determinant and its repeated-root limit>.
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