转发几个数列难题

realnumber

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踏歌而来

安徽2014年高考（21）题的（Ⅱ）也很难。

其妙

回复 2# 踏歌而来
广东2014年高考数列解答题难不难？

踏歌而来

回复 3# 其妙

这道题的（2）求数列的通项公式有点难。
直接 构造数列比较难，如果用猜想+数学归纳法则比较容易。

tommywong

16.

$\displaystyle a_n=(\frac{3+\sqrt{5}}{2})^n-2+(\frac{3-\sqrt{5}}{2})^n=[(\frac{\sqrt{5}+1}{2})^n-(\frac{\sqrt{5}-1}{2})^n]^2$

$\displaystyle b_n=(\frac{\sqrt{5}+1}{2})^n-(\frac{\sqrt{5}-1}{2})^n=1,\sqrt{5},4,3\sqrt{5},11,8\sqrt{5},...,b_{n+4}=3b_{n+2}-b_n$

tommywong

15.

$\displaystyle a_n=\frac{p_n}{q_n},a_{n+1}=\frac{p_n^2+\frac{q_n^2}{2}}{2p_n q_n}$

$n \neq 1$：

$\displaystyle p_{n+1}=p_n^2+\frac{q_n^2}{2},q_{n+1}=2p_n q_n$

$\displaystyle \sqrt{\frac{2}{2a_n^2-1}}=\frac{4a_{n-1}}{2a_{n-1}^2-1}=\frac{4p_{n-1}q_{n-1}}{2p_{n-1}^2-q_{n-1}^2}=\frac{q_n}{p_{n-1}^2-\frac{q_{n-1}^2}{2}}$

$\displaystyle p_{n-1}^2-\frac{q_{n-1}^2}{2}=(p_{n-2}^2+\frac{q_{n-2}^2}{2})^2-2p_{n-2}^2 q_{n-2}^2=(p_{n-2}^2-\frac{q_{n-2}^2}{2})^2=1$

$\displaystyle \frac{q_n}{p_{n-1}^2-\frac{q_{n-1}^2}{2}}=q_n$

tommywong

19.（f的无穷映射）

$2^{n_2} 3^{n_3} k \rightarrow 3^{n_3} k$

$n_3 \ge 1$时，$3^{n_3} k \rightarrow 3^{n_3} k+3 \rightarrow 3(\frac{3^{n_3-1}k+1}{2}) = 3^{n_3'} k' \rightarrow 3,6$，$3^{n_3} k>3^{n_3'} k'$

$n_3=0$时，$k \rightarrow k+3 \rightarrow \frac{k+3}{2} = k' \rightarrow 1,2,4$，$k>k'$

tommywong

续15.

$p_{n+1}=2p_n^2-1,q_{n+1}=2p_n q_n$

$\displaystyle a_n=\frac{(\sqrt{2}+1)^{2^n}+1}{\sqrt{2}[(\sqrt{2}+1)^{2^n}-1]}$

tommywong

还是15.

$\displaystyle a_{n+1}-\frac{1}{\sqrt{2}}=\frac{a_n}{2}-\frac{1}{\sqrt{2}}-\frac{1}{4a_n}=\frac{1}{2a_n}(a_n-\frac{1}{\sqrt{2}})^2,b_n=a_n-\frac{1}{\sqrt{2}},b_1=\frac{\sqrt{2}-1}{\sqrt{2}}$

$\displaystyle b_{n+1}=\frac{b_n^2}{2(b_n+\frac{1}{\sqrt{2}})},c_n=\frac{1}{b_n},c_1=2+\sqrt{2}$

$\displaystyle c_{n+1}=\sqrt{2}(c_n^2+\sqrt{2}c_n)=\sqrt{2}(c_n+\sqrt{1}{\sqrt{2}})^2-\sqrt{1}{\sqrt{2}},d_n=c_n+\frac{1}{\sqrt{2}},d_1=\frac{4+3\sqrt{2}}{2}$

$d_{n+1}=\sqrt{2}d_n^2,e_n=\sqrt{2}d_n,e_1=3+2\sqrt{2}$

$\displaystyle  e_n=e_{n-1}^2=(3+2\sqrt{2})^{2^{n-1}},d_n=\frac{1}{\sqrt{2}}(\sqrt{2}+1)^{2^n},c_n=\frac{1}{\sqrt{2}}(\sqrt{2}+1)^{2^n}-\frac{1}{\sqrt{2}}$

$\displaystyle b_n=\frac{\sqrt{2}}{(\sqrt{2}+1)^{2^n}-1},a_n=\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{(\sqrt{2}+1)^{2^n}-1}$

$\displaystyle \frac{4a_{n-1}}{2a_{n-1}^2-1}=\frac{1}{\sqrt{2}}[(\sqrt{2}+1)^{2^{n-1}}-(\sqrt{2}-1)^{2^{n-1}}]$

$x_n=(\sqrt{2}+1)^n-(\sqrt{2}-1)^n,x_{n+4}=6x_{n+2}-x_n$

tommywong

17.

bbs.emath.ac.cn/thread-5732-1-1.html
参见4#

tommywong

18.

设$x_n=\{2^n \sqrt{2}\}$为$2^n \sqrt{2}$的小数部分，
若$a_n$为奇数，$x_n<0.5$，则$a_{n+1}$为偶数。
若$a_n$为奇数，$x_n \ge 0.5$，则$a_{n+1}$为奇数，$x_{n+1}=2x_n-1=x_n-(1-x_n)$递减，当$x_{n+k}<0.5$时$a_{n+k+1}$为偶数。

$[2^n \sqrt{2}]$有无穷多个偶数