利用等差中项证明等差数列例题

（1）由题意知 $2 S_n-n a_n=3 n\left(n \in \mathbb{N}^*\right)$ ，

则 $2 S_{n+1}-(n+1) a_{n+1}=3(n+1)\left(n \in \mathbb{N}^*\right)$，

两式相减得: $2 S_{n+1}-2 S_n-(n+1) a_{n+1}+n a_n=3\left(n \in \mathbb{N}^*\right)$, 化简得: $n a_n=(n-1) a_{n+1}+3\left(n \in \mathbf{N}^*\right)$，

所以 $(n+1) a_{n+1}=n a_{n+2}+3\left(n \in \mathbf{N}^*\right)$，

两式相减得：$(n+1) a_{n+1}-n a_n=n a_{n+2}-(n-1) a_{n+1}\left(n \in \mathbb{N}^*\right)$，

化简得：$2 a_{n+1}=a_n+a_{n+2}\left(n \in \mathbf{N}^*\right)$ ，

令 $n=1$ ，得：$2 S_1-a_1=3$ ，

即 $2 a_1-a_1=3$ ，解得 $a_1=3$ ，

且 $a_2-a_1=5-3=2$ ， 所以数列 $\left\{a_n\right\}$ 为 3 为首项， 2 为公差的等差数列，

所以 $a_n=a_1+(n-1) d=3+2(n-1)=2 n+1 \left( n \in \mathbb{N}^* \right)$.

(2) 又(1)知 $a_n=2 n+1\left(n \in \mathbf{N}^*\right)$ ，

所以 $b_n=\dfrac{1}{a_n \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_n}}=\dfrac{1}{(2 n+1) \sqrt{2 n+3}+(2 n+3) \sqrt{2 n+1}}$

$=\dfrac{1}{\sqrt{2 n+1} \sqrt{2 n+3}(\sqrt{2 n+1}+\sqrt{2 n+3})}$

$=\dfrac{\sqrt{2 n+3}-\sqrt{2 n+1}}{\sqrt{2 n+1} \sqrt{2 n+3}(\sqrt{2 n+1}+\sqrt{2 n+3})(\sqrt{2 n+3}-\sqrt{2 n+1})} $

$=\dfrac{\sqrt{2 n+3}-\sqrt{2 n+1}}{2 \sqrt{2 n+1} \sqrt{2 n+3}} $

$=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{2 n+1}}-\dfrac{1}{\sqrt{2 n+3}}\right)\left(n \in \mathbb{N}^*\right)$

则 $T_n=b_1+b_2+\cdots+b_n$

$=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}\right)+\dfrac{1}{2}\left(\dfrac{1}{\sqrt{5}}-\dfrac{1}{\sqrt{7}}\right)+\cdots+\dfrac{1}{2}\left(\dfrac{1}{\sqrt{2 n+1}}-\dfrac{1}{\sqrt{2 n+3}}\right) $

$=\frac{1}{2}\left(\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{5}}-\dfrac{1}{\sqrt{7}}+\cdots+\dfrac{1}{\sqrt{2 n+1}}-\dfrac{1}{\sqrt{2 n+3}}\right) $

$=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{2 n+3}}\right)\left(n \in \mathbb{N}^*\right)$

令${T_n}>\dfrac{\sqrt{3}}{10}$，得$\dfrac{1}{2}\left(\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{2 n+3}}\right)>\dfrac{\sqrt{3}}{10}$ .

解得: $n>\dfrac{63}{8}$.

又 $n \in \mathbb{N}^*$ ，所以 $(n)_{\min}=8$.

所以使 $T_n>\dfrac{\sqrt{3}}{10}$ 成立的最小正整数 $n$ 的值为 8 .