证明:若$a_1\neq 0$,则
\begin{equation}\label{eq:2367} \frac{p_n}{p_{n-1}}=[a_n,a_{n-1},a_{n-2},\cdots,a_1]\end{equation}和\begin{equation}\label{eq:9876} \frac{q_n}{q_{n-1}}=[a_n,a_{n-1},a_{n-2},\cdots,a_2]\end{equation}
证明:我们先证明\ref{eq:2367}.真是神奇,竟然都倒过来了.由递推公式,当$n\geq 3$时,
\begin{equation}
p_n=a_np_{n-1}+p_{n-2}\end{equation}因此
\begin{equation} \frac{p_n}{p_{n-1}}=a_n+\frac{p_{n-2}}{p_{n-1}}\end{equation}设$\frac{p_n}{p_{n-1}}=k_n(n\geq 2)$,因此
\begin{equation} k_n=a_n+\frac{1}{k_{n-1}}\end{equation}这样就证明了$k_n=[a_n,a_{n-1},a_{n-2},\cdots,a_1]$了(为什么?). 下面来证明\ref{eq:9876}.根据递推公式,当$n\geq 3$时,\begin{equation} q_n=a_nq_{n-1}+q_{n-2}\end{equation}因此\begin{equation} \frac{q_n}{q_{n-1}}=a_n+\frac{q_{n-2}}{q_{n-1}}\end{equation}设$\frac{q_n}{q_{n-1}}=t_n(n\geq 2)$,因此\begin{equation} t_n=a_n+\frac{1}{t_{n-1}}\end{equation}因此也有$t_n=[a_n,a_{n-1},a_{n-2},\cdots,a_2]$.