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

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 $m$,

To form the next expression, first expand $4f(k)$:

Hence

Using the induction assumption,

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,

$4^n+6n-10$ is divisible by $18$ for all integers $n\geq2$.

比较相邻两项 $f(k+1)$ 与 $f(k)$。将差式改写为 $3f(k)$ 加上 $18$ 的整数倍，就能利用归纳假设完成证明。

Let $f(n)=4^n+6n-10$. When $n=2$,

so the result is true for $n=2$.

Assume that $f(k)=18m$ for some integer $m$, where $k\in\mathbb Z$ and $k\geq2$. Consider the difference:

Also,

Therefore

It follows that

Since $4m+2-k$ is an integer, $f(k+1)$ is divisible by $18$.

Thus the result for $n=k$ implies the result for $n=k+1$. By mathematical induction, $4^n+6n-10$ is divisible by $18$ for all integers $n\geq2$.