(i) 用数学归纳法证明：对于所有正整数 $n$，

(ii) 用数学归纳法证明：对于所有正整数 $n$， $f(n) = 8^{2n + 1} + 6^{2n - 1}$ 能被 $7$ 整除。

先验证 $n=1$。归纳步骤中，假设公式对 $n=k$ 成立，再把该矩阵的 $k$ 次幂右乘原矩阵。矩阵乘法后，每个元素都能提出因数 3，与外面的 $3^{k-1}$ 合并成 $3^k$。

Let

For $n=1$,

Hence 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 formula for $n=k+1$. Therefore, 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 $n\in\mathbb Z^+$.

先验证 $n=1$。归纳步骤中，把 $f(k+1)$ 凑成 $64f(k)$ 减去一个含因数 7 的项；利用归纳假设即可证明两部分都能被 7 整除。

For $n=1$,

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

Assume that the result is true for $n=k$. Thus

is divisible by 7.

Now,

By the induction hypothesis, $64f(k)$ is divisible by 7, and $28\big(6^{2k-1}\big)$ is divisible by 7. Hence $f(k+1)$ is divisible by 7.

Therefore, 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,

官方评分资料也允许改为凑出 $36f(k)$。此时剩下的是 $28\cdot8^{2k+1}$，同样显然含有因数 7。

For $n=1$,

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

Assume that $f(k)=8^{2k+1}+6^{2k-1}$ is divisible by 7. Then

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