(i) 用数学归纳法证明，对于所有正整数 $n \in \mathbb{Z}^+$ $\begin{pmatrix} -2 & 9 \\ -1 & 4 \end{pmatrix}^n = \begin{pmatrix} 1 - 3n & 9n \\ -n & 3n + 1 \end{pmatrix}$

(ii) 一个数列定义如下： $u_1 = 1 \quad u_2 = 4 \quad u_{n+2} = 6u_{n+1} - 9u_n$

用数学归纳法证明，对于所有正整数 $n \in \mathbb{Z}^+$， $u_n = 3^{n-2}(n + 2)$

先验证 $n=1$。再假设公式对 $n=k$ 成立，把假设中的矩阵右乘原矩阵，逐项计算并整理成把 $k$ 换成 $k+1$ 后的目标矩阵，最后写出完整归纳结论。

Let

For $n=1$, the right-hand side is

Hence the result is true for $n=1$.

Assume that the result is true for $n=k$, so that

Then

Therefore, if the result is true for $n=k$, it is also true for $n=k+1$. Since it is true for $n=1$, by mathematical induction,

for all $n\in\mathbb Z^+$.

这是二阶递推数列，归纳时必须验证 $n=1,2$ 两个初值，并同时假设公式对相邻的 $k$、$k+1$ 成立。把这两个假设代入递推式后，整理成公式在 $n=k+2$ 时的形式。

For $n=1$,

For $n=2$,

Hence the result is true for $n=1$ and $n=2$.

Assume that the result is true for $n=k$ and $n=k+1$, so that

and

Using the recurrence relation,

Therefore, if the result is true for $n=k$ and $n=k+1$, it is also true for $n=k+2$. Since it is true for $n=1$ and $n=2$, by mathematical induction,

for all $n\in\mathbb Z^+$.