18. (2013 广东) 设各项均为正数的数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 满足 $4 S_{n}=a_{n+1}^{2}-4 n-1, n \in N^{*}$,且 $a_{2}, a_{5}, a_{14}$ 构成等比数列.

( I ) 证明: $a_{2}=\sqrt{4 a_{1}+5}$;

(II) 求数列 $\left\{a_{n}\right\}$ 的通项公式;

(III) 证明: 对一切正整数 $n$, 有 $\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}<\frac{1}{2}$.
( I ) 当 $n=1$ 时, $4 a_{1}=a_{2}^{2}-5, a_{2}^{2}=4 a_{1}+5, \because a_{n}>0 \therefore a_{2}=\sqrt{4 a_{1}+5}$

(II) 当 $n \geq 2$ 时, $4 S_{n-1}=a_{n}^{2}-4(n-1)-1,4 a_{n}=4 S_{n}-4 S_{n-1}=a_{n+1}^{2}-a_{n}^{2}-4$

$a_{n+1}^{2}=a_{n}^{2}+4 a_{n}+4=\left(a_{n}+2\right)^{2}, \because a_{n}>0 \therefore a_{n+1}=a_{n}+2$

$\therefore$ 当 $n \geq 2$ 时, $\left\{a_{n}\right\}$ 是公差 $d=2$ 的等差数列.

$\because a_{2}, a_{5}, a_{14}$ 构成等比数列, $\therefore a_{5}^{2}=a_{2} \cdot a_{14},\left(a_{2}+8\right)^{2}=a_{2} \cdot\left(a_{2}+24\right)$, 解得 $a_{2}=3$,

由 ( I ) 可知, $4 a_{1}=a_{2}^{2}-5=4, \therefore a_{1}=1$

$\because a_{2}-a_{1}=3-1=2 \therefore\left\{a_{n}\right\}$ 是首项 $a_{1}=1$, 公差 $d=2$ 的等差数列.

$\therefore$ 数列 $\left\{a_{n}\right\}$ 的通项公式为 $a_{n}=2 n-1$.

(III) $\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}=\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\cdots+\frac{1}{(2 n-1)(2 n+1)}$

$=\frac{1}{2} \cdot\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{2 n-1}-\frac{1}{2 n+1}\right)\right]$

$=\frac{1}{2} \cdot\left[1-\frac{1}{2 n+1}\right]<\frac{1}{2}$.