数列题征解（1-5）

等待hxh

（一堆数列综合题，可惜没答案）

tommywong

2.
$\displaystyle a_{n+2}=\frac{(2a_{n+1}+3)(a_{n+1}-3)}{2a_n}$

$a_{n+1}=2a_n+3 \rightarrow a_{n+2}=2a_{n+1}+3$

tommywong

3.

$\displaystyle a_0+a_1+...+a_k+a_{k+1}=\frac{b_k(3^{2^{k+1}}+1)+2^k(3^{2^{k+1}}-1)}{(3^{2^{0}}+1)(3^{2^{1}}+1)...(3^{2^{k+1}}+1)}=\frac{b_{k+1}}{(3^{2^{0}}+1)(3^{2^{1}}+1)...(3^{2^{k+1}}+1)}$

$\displaystyle b_k=\frac{1}{4}(3^{2^{k+1}}-1)-2^k,A=\frac{2b_9}{3^{2^{10}}-1},B=\frac{2^{46}}{3^{2^{10}}-1}$

$\displaystyle \frac{A}{B}=\frac{3^{2^{10}}-2^{11}-1}{2^{47}}$

tommywong

4.

docin.com/p-76847173.html
第8页

tommywong

1.

$\displaystyle c_n=\frac{1}{a_n+1},d_n=\frac{1}{b_n+1},c_1=\frac{1}{2},d_1=\frac{1}{3}$

$\displaystyle c_{n+1}=c_n(1-d_n),d_{n+1}=d_n(1-c_n),c_{n+1}-d_{n+1}=c_n-d_n=\frac{1}{6}$

$\displaystyle c_{n+1}=c_n(\frac{7}{6}-c_n),d_{n+1}=d_n(\frac{5}{6}-d_n)$

$\displaystyle c_{\infty} \rightarrow \frac{1}{6},d_{\infty} \rightarrow 0$

$\displaystyle a_{\infty} \rightarrow 5,b_{\infty} \rightarrow \infty$

tommywong

5.

$x_{3n-2},x_{3n-1},x_{3n}=a_{3n-2},c_{3n-1},b_{3n}$

$y_{3n-2},y_{3n-1},y_{3n}=b_{3n-2},a_{3n-1},c_{3n}$

$z_{3n-2},z_{3n-1},z_{3n}=c_{3n-2},b_{3n-1},a_{3n}$

$\displaystyle x_1=1,y_1=2,z_1=3,x_{n+1}=x_n+\frac{1}{y_n},y_{n+1}=y_n+\frac{1}{z_n},z_{n+1}=z_n+\frac{1}{x_n}>z_n+\frac{1}{z_n}$

$\displaystyle z_n \ge \sqrt{an+b} \Leftrightarrow \sqrt{an+b}+\sqrt{\frac{1}{an+b}} > \sqrt{a(n+1)+b} \Leftrightarrow \frac{1}{an+b}+2 > a$

$max\{a_n,b_n,c_n\}=z_n \ge \sqrt{2n+7}$

其妙

回复 6# tommywong

caijinzhi

回复 5# tommywong

第一步的转化 请问您是怎么想到的？

tommywong

$\displaystyle b_{n+1}+1=(1+\frac{1}{a_n})(b_n+1),a_{n+1}+1=(1+\frac{1}{b_n})(a_n+1)$
$x_n=a_n+1,y_n=b_n+1$
$\displaystyle y_{n+1}=\frac{x_ny_n}{x_n-1},x_{n+1}=\frac{x_ny_n}{y_n-1}$
$\displaystyle c_n=\frac{1}{x_n},d_n=\frac{1}{y_n}$

caijinzhi

回复 9# tommywong
敬佩！