Re: [中學] 遞迴關係？
※ 引述《joey1226 (QQ)》之銘言：
: 經S大提醒已更正，不過因為在外頭僅能手機拍版，再次感謝
(a_{n+1}-5a_n)^2=24a_n^2+1
<=> a_{n+1}^2+a_n^2-10a_na_{n+1}=1
this holds for all n.
Thus, a_{n+1}^2+a_n^2-10a_na_{n+1}=a_n^2+a_{n-1}^2-10a_{n-1}a_n
i.e. a_{n+1}^2-a_{n-1}^2=10a_n(a_{n+1}-a_{n-1})
But a_{n+1}>a_{n-1}, so a_{n+1}+a_{n-1}=10a_n.
Hence a_{n+1}=10a_n-a_{n-1}.
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