9. 用数学归纳法证明，对于 $n \in \mathbb{Z}$，$n \geq 2$

$4^n + 6^n - 10$

能被 $18$ 整除。

先验证起始值 $n=2$。归纳步骤中，把“能被 $18$ 整除”写成 $4^k+6k-10=18m$，再将 $n=k+1$ 时的式子整理为 $18$ 与整数的乘积。

Let

For $n=2$,

which is divisible by $18$. Hence the result is true for $n=2$.

Assume that the result is true for $n=k$, where $k\in\mathbb Z$ and $k\geq2$. Then, for some integer $m$,

For $n=k+1$,

Since $m,k\in\mathbb Z$, $4m+2-k\in\mathbb Z$. Therefore $f(k+1)$ is divisible by $18$.

Thus, if the result is true for $n=k$, it is true for $n=k+1$. Hence, by mathematical induction,

for all integers $n\geq2$.