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[数列] 问个数列通项公式

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青青子衿 posted 2014-1-27 20:58 |Read mode
\[a_n=\frac{a_{n-1}+a_{n-2}}{1-a_{n-1}a_{n-2}}\]
有简单的通项公式吗??

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kuing posted 2014-1-27 21:07
你都照着 tan(x+y) 的公式来写递推了,还有什么问题?

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其妙 posted 2014-1-27 21:25
楼主这是相邻3项的递推关系,不是相邻3项的递推关系
自己随便想的题目吧?
original poster 青青子衿 posted 2014-1-29 16:10
回复 3# 其妙
楼主这是相邻3项的递推关系,不是相邻3项的递推关系
自己随便想的题目吧? ...
其妙 发表于 2014-1-27 21:25

\[a_n=\frac{a_{n-1}+a_{n-2}}{1-a_{n-1}a_{n-2}}\]\[a_1=tan \theta_1\]\[a_2=tan \theta_2\]\[a_3=tan (\theta_1+\theta_2)=tan \theta_3\]\[a_4=tan (\theta_2+\theta_3)=tan \theta_4\]\[a_5=tan (\theta_3+\theta_4)=tan \theta_5\]\[……\]
\[\theta_1+\theta_2=\theta_3\]\[\theta_2+\theta_3=\theta_4\]\[\theta_3+\theta_4=\theta_5\]\[……\]
\[\theta_n=F_n\]
\[a_n=\tan(F_n)\]

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其妙 posted 2014-2-3 14:28
牛笔!
那你还问什么题啊?自己都会做了还是kuing给的提示?
original poster 青青子衿 posted 2014-2-3 18:33
回复 5# 其妙
牛笔!
那你还问什么题啊?自己都会做了还是kuing给的提示?
其妙 发表于 2014-2-3 14:28
谢谢kuing!没错是kuing给的提示呀!

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