8. 用数学归纳法证明 $4^{n+2} + 5^{2n+1}$ 对所有正整数 $n$ 都能被 $21$ 整除。

设 $f(n)=4^{n+2}+5^{2n+1}$。先验证 $n=1$ 时可被 $21$ 整除；再假设 $f(k)$ 可被 $21$ 整除，把 $f(k+1)$ 改写成 $4f(k)$ 加上一个显然含因子 $21$ 的项。

Let

For $n=1$,

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

Assume that the result is true for $n=k$, where $k$ is a positive integer. Thus $f(k)$ is divisible by $21$.

Now

By the induction hypothesis, $4f(k)$ is divisible by $21$, and $21\cdot5^{2k+1}$ is also divisible by $21$. Therefore, $f(k+1)$ is divisible by $21$.

Hence, if the result is true for $n=k$, it is true for $n=k+1$. Since it is true for $n=1$, $4^{n+2}+5^{2n+1}$ is divisible by $21$ for all positive integers $n$, by mathematical induction.

也可以先配出 $25f(k)$，再减去多出的项，得到

归纳假设保证第一项可被 $21$ 整除，第二项也含有因子 $21$，所以它们的差同样可被 $21$ 整除。

Let

For $n=1$,

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

Assume that $f(k)$ is divisible by $21$ for some positive integer $k$. Then

By the induction hypothesis, $25f(k)$ is divisible by $21$. The term $21\cdot4^{k+2}$ is also divisible by $21$, so $f(k+1)$ is divisible by $21$.

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 positive integers $n$.