9.

(i)

用数学归纳法证明，对于 $n \in \mathbb{Z}^+$，$f(n)$ 是 $9$ 的倍数。

(ii) 一个数列定义如下：

$n\in\mathbb Z^+$

用数学归纳法证明，对于 $n \in \mathbb{Z}^+$

先验证 $n=1$。归纳步骤中直接计算 $f(k+1)-f(k)$，并将差整理成显然含有因子 $9$ 的形式；再结合归纳假设推出 $f(k+1)$ 也是 $9$ 的倍数。

For $n=1$,

which is a multiple of $9$.

Assume that the result is true for $n=k$, where $k\in\mathbb{Z}^+$. Thus $f(k)$ is a multiple of $9$.

Now

Therefore, $f(k+1)-f(k)$ is a multiple of $9$. Since $f(k)$ is also a multiple of $9$ by the inductive hypothesis, $f(k+1)$ is a multiple of $9$.

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 a multiple of $9$ for every $n\in\mathbb{Z}^+$ by mathematical induction.

不先作差，而是直接把 $f(k+1)$ 拆成 $f(k)$ 与一个显然含因子 $9$ 的余项。

The base case is

which is a multiple of $9$.

Assume that $f(k)$ is a multiple of $9$ for some $k\in\mathbb{Z}^+$. Then

Both terms on the final line are multiples of $9$, so $f(k+1)$ is a multiple of $9$.

Therefore, by mathematical induction, $f(n)$ is a multiple of $9$ for every positive integer $n$.

把差分推广为先减去 $(9m+1)f(k)$。展开后其余各项都含有因子 $9$，于是任意固定的非负整数 $m$ 都能完成归纳步骤。

The base case is $f(1)=27$, a multiple of $9$.

Assume that $f(k)$ is a multiple of $9$ for a positive integer $k$. Fix a non-negative integer $m$, and define

Using the definition of $f$,

Write the coefficient of $7^k$ as $H$:

Thus, put

which is an integer. Then

So $D_m$ is a multiple of $9$. Since

the inductive hypothesis shows that $f(k+1)$ is also a multiple of $9$.

Together with the base case, this proves the statement for all positive integers $n$ by mathematical induction.

例如，$m=0$ 给出第一种差分；其他取值则有

这些都是同一个归纳步骤的变形。

若凑出 $7f(k)$，余项是 $g(k)=6+21\cdot7^k$。它不是一眼可见的 $9$ 的倍数，所以同时对 $f$ 与 $g$ 做归纳，先验证两个初值，再证明两者的下一项都能被 $9$ 整除。

Define

At $n=1$,

so both are multiples of $9$.

Assume that both $f(k)$ and $g(k)$ are multiples of $9$ for a positive integer $k$. First,

Hence $g(k+1)$ is a multiple of $9$. Also,

Both terms on the last line are multiples of $9$. Therefore $f(k+1)$ is a multiple of $9$ as well.

The simultaneous inductive step and the two base values prove that both $f(n)$ and $g(n)$ are multiples of $9$ for every positive integer $n$.

辅助式也可以换成 $h(n)=3\cdot7^n-3$，因为

最后两式同样给出 $h(n)$ 能被 $9$ 整除的完整归纳步骤，再代回第一式即可。

这是二阶递推关系，因此归纳步骤需要同时假设公式对连续两项 $u_k$、$u_{k+1}$ 成立，才能推出 $u_{k+2}$。相应地，基础步骤也必须验证 $n=1$ 和 $n=2$ 两项。

For $n=1$,

For $n=2$,

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

Assume that the result is true for $n=k$ and $n=k+1$, where $k\in\mathbb{Z}^+$. Hence

and

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 $n=2$,

for every positive integer $n$, by mathematical induction.