(i) 数列 $u_1, u_2, u_3, \ldots$ 定义为 $u_{n+1} = \frac{1}{3}(2u_n - 1) \quad u_1 = 1$

用数学归纳法证明：对于 $n \in \mathbb{Z}^+$， $u_n = 3\left(\frac{2}{3}\right)^n - 1$

(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 $2f(k)+7\cdot3^{2k+1}$；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 $n\in\mathbb Z^+$.