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[数列] 递推,求通项。

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guanmo1 posted 2022-10-9 09:21 |Read mode
Last edited by hbghlyj 2025-3-19 08:30$a_{n+1}^2=a_n+2\,(0<a_1<2)$, 求 $a_n$.

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战巡 posted 2022-10-9 10:58
\[a_{n}=2\cos(\arccos(\frac{a_1}{2})\cdot 2^{-n+1})\]

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original poster guanmo1 posted 2022-10-9 11:46
战巡 发表于 2022-10-9 10:58
\[a_{n}=2\cos(\arccos(\frac{a_1}{2})\cdot 2^{-n+1})\]
过程呢,谢谢。

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kuing posted 2022-10-9 13:59
应当限制 `a_n` 为正,否则 `a_2` 有两个值,`a_3` 有四个值。

然后令 `a_n=2\cos b_n`(`b_n\in(0,\pi/2)`)即可。

简短的题目还是请你尝试码字。

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战巡 posted 2022-10-9 21:26
guanmo1 发表于 2022-10-9 11:46
过程呢,谢谢。
没过程,一眼看出的

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original poster guanmo1 posted 2022-10-9 22:55 from mobile
战巡 发表于 2022-10-9 21:26
没过程,一眼看出的
现在知道套路了,谢谢!

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