9. (i) 一个数列定义如下：

$u_1 = 0 \quad u_2 = -6$

$u_{n+2} = 5u_{n+1} - 6u_n \quad n \geq 1$

用数学归纳法证明，对于 $n \in \mathbb{Z}^+$

$u_n = 3 \times 2^n - 2 \times 3^n$

(ii) 用数学归纳法证明，对于所有正整数 $n$

$f(n) = 3^{3n-2} + 2^{4n-1}$

能被 $11$ 整除。

递推关系同时依赖前两项，因此归纳基础必须验证 $n=1$ 和 $n=2$。归纳步骤中假设公式对 $n=k$、$n=k+1$ 都成立，将两式代入递推关系，证明公式对 $n=k+2$ 成立。

For $n=1$, the proposed formula gives

For $n=2$, it gives

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

Assume that the result is true for $n=k$ and $n=k+1$. Then

Using the recurrence relation,

Therefore, if the result is true for $n=k$ and $n=k+1$, it is true for $n=k+2$. Since it is true for $n=1$ and $n=2$, the result is true for all $n\in\mathbb{Z}^+$ by mathematical induction.

先验证 $n=1$。假设 $f(k)$ 可被 $11$ 整除，把 $f(k+1)$ 中的指数降到与 $f(k)$ 一致，再整理成 $16f(k)$ 与一个显然含因子 $11$ 的项之和。

For $n=1$,

which is divisible by $11$.

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

is divisible by $11$.

Now

By the induction hypothesis, $16f(k)$ is divisible by $11$, and $11\times3^{3k-2}$ is also divisible by $11$. Therefore, $f(k+1)$ is divisible by $11$.

Hence, if the result is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=1$, $f(n)$ is divisible by $11$ for all positive integers $n$, by mathematical induction.