9. (i) 数列 $u_1, u_2, u_3, \ldots$ 定义为

用数学归纳法证明：对于 $n \in \mathbb{Z}^+$，

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

用数学归纳法证明：对于 $n \in \mathbb{Z}^+$，$f(n)$ 是 $7$ 的倍数。

先验证 $n=1$。归纳步骤假设公式对 $n=k$ 成立，把 $u_k$ 的公式代入递推关系，并整理成原公式中将 $n$ 替换为 $k+1$ 的形式。

For $n=1$, the proposed formula gives

Therefore, the result is true for $n=1$.

Assume that the result is true for $n=k$, where $k\in\mathbb Z^+$. Thus

Using the recurrence relation,

This is the required formula for $n=k+1$. Hence, if the result is true for $n=k$, it is true for $n=k+1$.

Since the result is true for $n=1$, by mathematical induction,

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

先验证 $f(1)$ 是 $7$ 的倍数。归纳步骤把 $f(k+1)$ 重组为

归纳假设保证第一项是 $7$ 的倍数，第二项显然也是。

For $n=1$,

Therefore, the result is true for $n=1$.

Assume that the result is true for $n=k$, where $k\in\mathbb Z^+$. Thus

is a multiple of $7$.

Now

By the induction hypothesis, $2f(k)$ is a multiple of $7$, and $7\cdot3^{2k+1}$ is also a multiple of $7$. Therefore, $f(k+1)$ is a multiple of $7$.

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

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

也可以把 $f(k+1)$ 重组为

这次先配出 $9f(k)$，再减去多出的 $7\cdot2^{k+2}$。两项都是 $7$ 的倍数，因而能完成归纳步骤。

For $n=1$,

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

Assume that $f(k)$ is a multiple of $7$ for some $k\in\mathbb Z^+$, where

Then

By the induction hypothesis, $9f(k)$ is a multiple of $7$. Also, $7\cdot2^{k+2}$ is a multiple of $7$, so their difference $f(k+1)$ is a multiple of $7$.

Hence the result being true for $n=k$ implies that it is true for $n=k+1$. Since it is true for $n=1$, mathematical induction gives

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