设递增数列 $\{a_n\}$ 满足 $a_1=1,4a_{n+1}=5a_n+\sqrt{9a^2_n+16}(n\geqslant1,n\in\mathbb N^*)$．
【难度】
【出处】
2012年全国高中数学联赛辽宁省预赛
【标注】
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求数列 $\{a_n\}$ 的通项公式；标注答案$a_n=\dfrac23\left(2^n-\dfrac{1}{2^n}\right)$解析由 $4a_{n+1}=5a_n+\sqrt{9a_n^2+16}$，得$$a_{n+1}^2+a_n^2-\dfrac52a_{n+1}a_n=1,$$故$$a_n^2+a_{n-1}^2-\dfrac52a_na_{n-1}=1(n>1),$$两式相减得$$(a_{n+1}-a_{n-1})\left(a_{n+1}-a_{n-1}-\dfrac52a_n\right)=0.$$又 $\{a_n\}$ 为递增数列，故$$a_{n+1}-\dfrac52a_n+a_{n-1}=0(n>1),$$其特征方程为 $x^2-\dfrac52x+1=0$，特征根为 $x_1=2,x_2=\dfrac12$，所以$$a_n=a_1x_1^n+a_2x_2^n.$$将 $a_1=1,a_2=\dfrac52$ 代入，得$$\begin{cases}2a_1+\dfrac12a_2=1,\\4a_1+\dfrac14a_2=\dfrac52,\end{cases}$$解得 $a_1=\dfrac23,a_2=-\dfrac23$．
因此通项公式为 $a_n=\dfrac23\left(2^n-\dfrac{1}{2^n}\right)$． -
证明：$\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_n}<2$．标注答案略解析设 $S_n=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_n}$．
因为$$2^n-\dfrac{1}{2^n}=2\left(2^{n-1}-\dfrac{1}{2^{n+1}}\right)>2\left(2^{n-1}-\dfrac{1}{2^{n-1}}\right)(n\geqslant2),$$所以$$a_n>2a_{n-1} , \dfrac{1}{a_n}<\dfrac12\cdot\dfrac{1}{a_{n-1}}(n\geqslant2),$$故\[\begin{split}S_n&=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_n}\\&<\dfrac{1}{a_1}+\dfrac12\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_{n-1}}\right)\\&=\dfrac{1}{a_1}+\dfrac12\left(S_n-\dfrac{1}{a_n}\right)\\&<\dfrac{1}{a_1}+\dfrac12\cdot S_n,\end{split}\]因此 $S_n<\dfrac{2}{a_1}=2$．
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