每日一题[2294]类等比数列

已知数列 $\left\{a_{n}\right\}$ 中，$a_{1}=\sqrt{2}$，$a_{2}=2$，$a_{n+2}=\dfrac{a_{n+1}^{2}+2^{n+2}}{a_{n}}$，$n \in \mathbb{N}^{*}$，求数列 $\left\{a_{n}\right\}$ 的通项公式．

答案    $a_n=\dfrac{1}{\sqrt 2}\left(\left(2\sqrt 2+\sqrt 6\right)^{n-1}+ \left(2\sqrt 2-\sqrt 6\right)^{n-1}\right)$（$n\in\mathbb N^{\ast}$）．

解析    根据题意，有 $a_3=6\sqrt 2$，当 $n\geqslant 2$ 时，有

$$a_{n+2}a_n-a_{n+1}^2=2\left(a_{n+1}a_{n-1}-a_n^2\right)\implies \dfrac{a_{n+2}+2a_n}{a_{n+1}}=\dfrac{a_{n+1}+2a_{n-1}}{a_n}=\cdots=\dfrac{a_3+2a_1}{a_2}=4\sqrt 2,$$ 因此 $$a_{n+2}=4\sqrt 2a_{n+1}-2a_n,$$ 该递推式对应的特征根为 $x=2\sqrt 2\pm\sqrt 6$，于是 $$a_n=A\cdot \left(2\sqrt 2+\sqrt 6\right)^{n-1}+B\cdot \left(2\sqrt 2-\sqrt 6\right)^{n-1},n\in\mathbb N^{\ast},$$ 其中待定系数 $A,B$ 满足 $$\begin{cases} A+B=\sqrt 2,\\ A\cdot \left(2\sqrt 2+\sqrt 6\right)+B\cdot \left(2\sqrt 2-\sqrt 6\right)=2,\end{cases}\iff \begin{cases} A=\dfrac1{\sqrt 2},\\ B=\dfrac1{\sqrt 2},\end{cases}$$ 因此 $a_n=\dfrac{1}{\sqrt 2}\left(\left(2\sqrt 2+\sqrt 6\right)^{n-1}+ \left(2\sqrt 2-\sqrt 6\right)^{n-1}\right)$（$n\in\mathbb N^{\ast}$）．