\[
\begin{split}
D_n(t)&=\begin{vmatrix}
0 & 1 & 1 & \cdots & 1\\
t & 0 & 1 & \cdots & 1\\
t & t & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \ddots & 1\\
t & t & t & \cdots & 0\\
\end{vmatrix}_n\\
&\xlongequal[i=n,n-1,\ldots,2]{r_i-r_{i-1}}
\begin{vmatrix}
0 & 1 & 1 & \cdots & 1\\
t & -1 & 0 & \cdots & 0\\
0 & t & -1 & \cdots & 0\\
\vdots & \vdots & \ddots & \ddots & 0\\
0 & 0 & \cdots & t & -1\\
\end{vmatrix}_n\\
&\xlongequal[i=1,2,\ldots,n-1]{c_i+c_{i-1}\cdot\frac{1}{t}}
\begin{vmatrix}
0 & 1 & 1+\frac{1}{t} & \cdots & 1+\frac{1}{t}+\cdots+\frac{1}{t^{n-2}}\\
t & 0 & 0 & \cdots & 0\\
0 & t & 0 & \cdots & 0\\
\vdots & \vdots & \ddots & \ddots & 0\\
0 & 0 & \cdots & t & 0\\
\end{vmatrix}_n\\
&=(-1)^{1+n}(1+\frac{1}{t}+\cdots+\frac{1}{t^{n-2}})\cdot t^{n-1}\\
&=(-1)^{n-1}(t+t^2+t^3+\cdots+t^{n-1}).
\end{split}
\]
若 $t=1$, 则 $D_n(t)=(-1)^{n-1}\cdot(n-1)$. 从而原行列式为
\[
a_n\cdot D_n(t)=(-1)^{n-1}\cdot(n-1)a^n.
\]
若 $t\neq 1$, 则原式等于
\[
\begin{split}
a^n D_n(t)&=a^n\cdot(-1)^{n-1}(t+t^2+t^3+\cdots+t^{n-1})\\
&=a^n\cdot(-1)^{n-1}\cdot t(1+t+t^2+\cdots+t^{n-2})\\
&=a^n\cdot(-1)^{n-1}\cdot t\cdot\frac{1-t^{n-1}}{1-t}\\
&=a^n\cdot(-1)^{n-1}\cdot\frac{b}{a}\cdot\frac{1-(\frac{b}{a})^{n-1}}{1-\frac{b}{a}}\\
&=(-1)^{n-1}ab\frac{a^{n-1}-b^{n-1}}{a-b}.
\end{split}
\]
注意到
\[
\tilde{D}_n=
\begin{vmatrix}
a+b & ab & 0 & 0 & \cdots & 0\\
1 &a+b &ab & 0 & \cdots & 0\\
0 & 1 &a+b &ab & \cdots & 0\\
\vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\
0 & 0 & \cdots & 1 & a+b & ab \\
0 & 0 & 0 & \cdots & 1 & a+b \\
\end{vmatrix}=\left\{
\begin{aligned}
(n+1)a^n, \quad a=b,\\
\frac{a^{n+1}-b^{n+1}}{a-b}, \quad a\neq b.
\end{aligned}\right.
\]
因此我们有
\[
D_{n+2}=(-1)^{n+1}\cdot ab\cdot\tilde{D}_n.
\]
