8. (i) 用数学归纳法证明，对于 $n \in \mathbb{Z}^+$

$\sum_{r=1}^{n} \frac{2r^2 - 1}{r^2(r+1)^2} = \frac{n^2}{(n+1)^2}$

(ii) 用数学归纳法证明，对于 $n \in \mathbb{Z}^+$

$f(n) = 12^n + 2 \times 5^{n-1}$

能被 $7$ 整除。

先验证 $n=1$。归纳步骤中假设结论对 $n=k$ 成立，再给等式左边加上第 $k+1$ 项；通分后把分子展开成 $(k+1)^4$，即可得到 $n=k+1$ 时的目标式。

For $n=1$,

and

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

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

Then

This is the required result with $k+1$ in place of $n$. Hence, if the result is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=1$, the result is true for all $n\in\mathbb{Z}^+$ by mathematical induction.

先验证 $f(1)$ 能被 $7$ 整除。归纳步骤中把 $f(k+1)$ 写成 $12f(k)-14\cdot5^{k-1}$；归纳假设保证第一项能被 $7$ 整除，第二项显然也含因子 $7$。

For $n=1$,

which is divisible by $7$.

Assume that the result is true for $n=k$, where $k\in\mathbb{Z}^+$. Thus $f(k)$ is divisible by $7$.

Now

By the inductive hypothesis, $12f(k)$ is divisible by $7$, and $14\times5^{k-1}$ is also divisible by $7$. Therefore, $f(k+1)$ is divisible by $7$.

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 $7$ for all $n\in\mathbb{Z}^+$ by mathematical induction.

归纳框架不变，但在归纳步骤中计算 $f(k+1)-5f(k)$。含 $5^k$ 的项恰好相消，剩下 $7\cdot12^k$，因而整除关系更直接。

For $n=1$,

which is divisible by $7$.

Assume that $f(k)$ is divisible by $7$ for some $k\in\mathbb{Z}^+$. Then

Therefore,

Both terms on the right are divisible by $7$, so $f(k+1)$ is divisible by $7$.

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 $7$ for all $n\in\mathbb{Z}^+$ by mathematical induction.