解答题 （共 10 题 ），解答过程应写出必要的文字说明、证明过程或演算步骤
试证
$$
D_n=\left|\begin{array}{cccccc}
\alpha+\beta & \alpha \beta & 0 & \cdots & 0 & 0 \\
1 & \alpha+\beta & \alpha \beta & \cdots & 0 & 0 \\
0 & 1 & \alpha+\beta & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & \alpha+\beta & \alpha \beta \\
0 & 0 & 0 & \cdots & 1 & \alpha+\beta
\end{array}\right|=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}(\alpha \neq \beta) .
$$
试证
$$
D_n=\left|\begin{array}{cccccc}
x & -1 & 0 & \cdots & 0 & 0 \\
0 & x & -1 & \cdots & 0 & 0 \\
0 & 0 & x & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & x & -1 \\
a_n & a_{n-1} & a_{n-2} & \cdots & a_2 & a_1+x
\end{array}\right| \text {. } \begin{gathered}
\\
=x^n+a_1 x^{n-1}+a_2 x^{n-2}+\cdots+a_{n-1} x+a_n \quad(n \geqslant 2) .
\end{gathered}
$$
计算$D_n \xlongequal{ }\left|\begin{array}{cccccc}
a & -1 & 0 & \cdots & 0 & 0 \\
a x & a & -1 & \cdots & 0 & 0 \\
a x^2 & a x & a & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
a x^{n-2} & a x^{n-3} & a x^{n-4} & \cdots & a & -1 \\
a x^{n-1} & a x^{n-2} & a x^{n-3} & \cdots & a x & a
\end{array}\right| .$
计算$D_n=\left|\begin{array}{cccccc}
x & -1 & 0 & \cdots & 0 & 0 \\
0 & x & -1 & \cdots & 0 & 0 \\
0 & 0 & x & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & x & -1 \\
a_n & a_{n-1} & a_{n-2} & \cdots & a_2 & a_1+x
\end{array}\right| $
计算$D_n=\left|\begin{array}{cccccc}
\alpha+\beta & \alpha \beta & 0 & \cdots & 0 & 0 \\
1 & \alpha+\beta & \alpha \beta & \cdots & 0 & 0 \\
0 & 1 & \alpha+\beta & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & \alpha+\beta & \alpha \beta \\
0 & 0 & 0 & \cdots & 1 & \alpha+\beta
\end{array}\right| .$
计算$D_n=\left|\begin{array}{cccccc}
x & -1 & 0 & \cdots & 0 & 0 \\
0 & x & -1 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & x & -1 \\
a_n & a_{n-1} & a_{n-2} & \cdots & a_2 & a_1+x
\end{array}\right|(x \neq 0)$
设4阶行列式
$$
D_4 \xlongequal{ }\left|\begin{array}{cccc}
1 & -1 & 0 & -1 \\
-1 & 0 & 2 & 1 \\
1 & 1 & -1 & -2 \\
3 & 0 & 1 & 2
\end{array}\right|
$$
求：$A_{11}+A_{12}-A_{13}-2 A_{14}$（ $A_{i j}$ 是元素 $a_{i j}$ 的代数余子式）．
设4阶行列式
$$
D=\left|\begin{array}{cccc}
1 & 0 & -1 & 2 \\
1 & 1 & 2 & 1 \\
-2 & 1 & 0 & 2 \\
1 & -1 & 1 & 1
\end{array}\right|
$$
已知 $n$ 阶行列式
$$
D_n=\left|\begin{array}{ccccc}
2 & 4 & 6 & \cdots & 2 n \\
1 & 2 & & & \\
1 & & 3 & & \\
\vdots & & & \ddots & \\
1 & & & & n
\end{array}\right|
$$
求：$S=A_{11}+A_{12}+\cdots+A_{1 n}$（不计算 $A_{1 j}$ 求 $S$ ）．
已知5阶行列式
$$
D_5=\left|\begin{array}{lllll}
1 & 2 & 1 & 2 & 3 \\
2 & 2 & 2 & 1 & 1 \\
3 & 1 & 2 & 4 & 5 \\
1 & 1 & 1 & 2 & 2 \\
2 & 3 & 1 & 0 & 1
\end{array}\right|=-9
$$
求：$S_1=A_{41}+A_{42}+A_{43}$ 及 $S_2=A_{44}+A_{45}$ ．