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[数列] 一道数列题的第三问

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踏歌而来 posted 2015-6-24 10:25 |Read mode
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第一、第二问都没有问题。
第三问不知如何解答。
请大师们帮看看!

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kuing posted 2015-6-24 12:19
凡是问第N问的,请将前N-1问的答案给出。
&活着存在 posted 2015-6-24 15:02
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战巡 posted 2015-6-24 17:01
回复 1# 踏歌而来


\[a_n=2^n+(n-1)\lambda^n\]
\[\frac{a_2}{a_1}=\frac{1}{2}(4+\lambda^2)\]
\[\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\begin{cases} \lambda, \lambda\ge 2\\ 2, 0<\lambda<2\end{cases}\]
很容易证明不管什么情况,都有
\[\frac{a_2}{a_1}=\frac{1}{2}(4+\lambda^2)>\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\]
而这数列又没有$0$点,不存在突然跳起一个间断点的事情,显然$\frac{a_{n+1}}{a_n}$是有上界的,而且上界不会在极限处取到,因此得证
反正没说非要求出$k$,只要证明存在就够了

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original poster 踏歌而来 posted 2015-6-24 19:52
回复 4# 战巡

实际上,K就是1,用几何画板能看到。
上式只能说明n趋向于+∞时,成立。
如何能够证明n取任何一个正整数都小于等于a2/a1呢?

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original poster 踏歌而来 posted 2015-6-24 19:54
回复 2# kuing

刚刚看到。
3楼已经给出了通项公式。

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战巡 posted 2015-6-24 20:14
回复 5# 踏歌而来


都说了只需要证明$k$存在就可以了,没必要多此一举去证什么$k=1$

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original poster 踏歌而来 posted 2015-6-24 21:20
回复 7# 战巡

a2/a1对应的就是k=1。

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original poster 踏歌而来 posted 2015-6-24 21:48

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original poster 踏歌而来 posted 2015-6-26 16:39
Last edited by 踏歌而来 2015-6-26 20:17现在找到了k=1的证明方法。
谢谢大家!

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其妙 posted 2015-6-27 00:00
尼玛这不是天津高考题么?

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