已知数列 $a_1,a_2,\cdots,a_n,\cdots$ 满足：$a_1=1,a_2=1,a_{n+1}=\dfrac {n^2a_n^2+5}{(n^2-1)a_{n-1}}(n\geqslant 2)$，求 $a_n$ 的通项公式．
【难度】
【出处】
2008年全国高中数学联赛天津市预赛
【标注】
【答案】
$a_n =\dfrac 1n\left(\dfrac {63-13\sqrt {21}}{42}\left(\dfrac {5+\sqrt {21}}{2}\right)^n+\dfrac {63+13\sqrt {21}}{42}\left(\dfrac {5-\sqrt {21}}{2}\right)^n\right)$
【解析】
由已知得\[\begin{split}(n^2-1)a_{n-1}a_{n+1}&=n^2a_n^2+5\\((n-1)^2-1)a_{n-2}a_n&=(n-1)^2a_{n-1}^2+5,\end{split}\]两式相减得$$(n^2-1)a_{n-1}a_{n+1}-n(n-2)a_{n-2}a_n=n^2a_n^2-(n-1)^2a_{n-1}^2,$$即$$(n-1)a_{n-1}((n+1)a_{n+1}+(n-1)a_{n_1})=na_n(na_n+(n-2)a_{n-2}).$$设 $b_n=na_n(n\geqslant 1)$，则有\[\begin{split}b_{n-1}(b_{n+1}+b_{n-1})&=b_n(b_n+b_{n-2}),\\ \dfrac {b_{n+1}+b_{n-1}}{b_n}&=\dfrac {b_n+b_{n-2}}{b_{n-1}}.\end{split}\]设 $c_n=\dfrac {b_n+b_{n-2}}{b_{n-1}}(n\geqslant 3)$，由 $a_1=1,a_2=1,a_3=3$，可得 $b_1=1,b_2=2,b_3=9$，于是有$$c_n=c_{n-1}=\cdots =c_3=\dfrac {b_3+b_1}{b_2}=5.$$因为 $b_n-5b_{n-1}+b_{n-2}=0$，所以特征方程为 $x^2-5x+1=0$，特征根为 $x_{1,2}=\dfrac {5\pm {\sqrt {21}}}2$，从而设$$b_n=\lambda _1\left(\dfrac {5+{\sqrt {21}}}2\right)^n+\lambda _2\left(\dfrac {5-\sqrt {21}}2\right)^n.$$由 $b_1=1,b_2=2$ 及 $b_n-5b_{n-1}+b_{n-2}=0$，定义 $b_0=3$，于是有\[\begin{split}3&=b_0=\lambda _1+\lambda _2,\\1&=b_1=\lambda _1\left(\dfrac {5+\sqrt {21}}2\right)+\lambda _2\left(\dfrac {5-\sqrt {21}}2\right).\end{split}\]从而可得$$\lambda _1=\dfrac {63-13\sqrt {21}}{42},\lambda _2=\dfrac {63+13\sqrt {21}}{42}.$$因此有\[\begin{split}b_n&=\dfrac {63-13\sqrt {21}}{42}\left(\dfrac {5+\sqrt {21}}{2}\right)^n+\dfrac {63+13\sqrt {21}}{42}\left(\dfrac {5-\sqrt {21}}{2}\right)^n,\\a_n&=\dfrac 1n\left(\dfrac {63-13\sqrt {21}}{42}\left(\dfrac {5+\sqrt {21}}{2}\right)^n+\dfrac {63+13\sqrt {21}}{42}\left(\dfrac {5-\sqrt {21}}{2}\right)^n\right).\end{split}\]
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