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标题: c(1)=5／3，c(n)=bn／an 为最简分数时 c(n+1)=(an+2)／(bn+1)，求 c(1)+c(2)+…+c(21) [打印本页]

作者: a0952775081 时间: 2023-11-28 09:18
标题: c(1)=5／3，c(n)=bn／an 为最简分数时 c(n+1)=(an+2)／(bn+1)，求 c(1)+c(2)+…+c(21)
請問這題

[attach]136329[/attach]

作者: 天山草 时间: 2023-11-28 10:25
本帖最后由天山草于 2023-11-28 10:42 编辑

已知数列 c(n) 的首项为 c(1) = 5/3; c(n+1) = a(n+2)/b(n+1);
a(0)=1; a(n)=a(n-1)+2; b(0)=1; b(n)=b(n-1)+1;
求 s(21)=c(1)+c(2)+c(3)+...+c(21) 化为最简分数的准确值。

a[0] = 1; a[n_] := a[n] = (a[n - 1] + 2);
b[0] = 1; b[n_] := b[n] = (b[n - 1] + 1);
c[n_] := c[n] = (a[n] + 2)/(b[n] + 1);
s[0] = 0; s[n_] := s[n] = (s[n - 1] + c[n]);
Do[ Print["b(", n, ") = ", b[n], ", a(", n, ") = ", a[n], ", c(", n, ") = ", c[n], ", s(", n, ") = ", s[n]], {n, 1, 21}];

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运行结果：
b(1) = 2,  a(1) = 3,  c(1) = 5/3,  s(1) = 5/3
b(2) = 3,  a(2) = 5,  c(2) = 7/4,  s(2) = 41/12
b(3) = 4,  a(3) = 7,  c(3) = 9/5,  s(3) = 313/60
b(4) = 5,  a(4) = 9,  c(4) = 11/6,  s(4) = 141/20
b(5) = 6,  a(5) = 11,  c(5) = 13/7,  s(5) = 1247/140
b(6) = 7,  a(6) = 13,  c(6) = 15/8,  s(6) = 3019/280
b(7) = 8,  a(7) = 15,  c(7) = 17/9,  s(7) = 31931/2520
b(8) = 9,  a(8) = 17,  c(8) = 19/10,  s(8) = 36719/2520
b(9) = 10,  a(9) = 19,  c(9) = 21/11,  s(9) = 456829/27720
b(10) = 11,  a(10) = 21,  c(10) = 23/12,  s(10) = 509959/27720
b(11) = 12,  a(11) = 23,  c(11) = 25/13,  s(11) = 7322467/360360
b(12) = 13,  a(12) = 25,  c(12) = 27/14,  s(12) = 8017447/360360
b(13) = 14,  a(13) = 27,  c(13) = 29/15,  s(13) = 8714143/360360
b(14) = 15,  a(14) = 29,  c(14) = 31/16,  s(14) = 18824681/720720
b(15) = 16,  a(15) = 31,  c(15) = 33/17,  s(15) = 343803337/12252240
b(16) = 17,  a(16) = 33,  c(16) = 35/18,  s(16) = 122542379/4084080
b(17) = 18,  a(17) = 35,  c(17) = 37/19,  s(17) = 2479416161/77597520
b(18) = 19,  a(18) = 37,  c(18) = 39/20,  s(18) = 526146265/15519504
b(19) = 20,  a(19) = 39,  c(19) = 41/21,  s(19) = 185482083/5173168
b(20) = 21,  a(20) = 41,  c(20) = 43/22,  s(20) = 195593275/5173168
b(21) = 22,  a(21) = 43,  c(21) = 45/23,  s(21) = 4731437885/118982864

作者: 天山草 时间: 2023-11-28 11:09
这道题如果不用编程计算，有别的巧妙方法吗？

作者: Treenewbee 时间: 2023-11-28 11:27
天山草的初始值不对，应该是：\[\frac{5}{3},\frac{5}{6},\frac{4}{3},1,\frac{3}{2},1,\frac{3}{2},1,\frac{3}{2},1,\frac{3}{2},1,\frac{3}{2},1,\frac{3}{2},1,\frac{3}{2},1,\frac{3}{2},1,\frac{3}{2}......\]

作者: Treenewbee 时间: 2023-11-28 11:55

天山草发表于 2023-11-28 03:09
这道题如果不用编程计算，有别的巧妙方法吗？

此类题目总是需要先列出前几项的，你列出三五项就会发现它进入循环了，如此就简单了

作者: Treenewbee 时间: 2023-11-28 11:57

天山草发表于 2023-11-28 02:25
已知数列 c(n) 的首项为 c(1) = 5/3; c(n+1) = a(n+2)/b(n+1);
a(0)=1; a(n)=a(n-1)+2; b(0)=1; b(n) ...

复杂了，可简化为

c=NestList[Simplify@(Denominator@#+2)/(Numerator@#+1)&,5/3,20];
{c,Total@c}

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作者: a0952775081 时间: 2023-11-28 12:49
謝謝各位老師，我有列出來以及算出答案

作者: 天山草 时间: 2023-11-29 09:19
本帖最后由天山草于 2023-11-29 09:49 编辑

按 4# 楼的说法，原题的意思是：

[attach]136365[/attach]

那么原题怎样表达才更清楚一点呢？ a（n），b（n）的递推式是怎样的？

作者: Treenewbee 时间: 2023-11-29 10:56
\[a_n=\{3,6,3,1,2,1,2,1,2,1....\}\]
\[b_n=\{5,5,4,1,3,1,3,1,3,1....\}\]

作者: Treenewbee 时间: 2023-11-29 17:42

Treenewbee 发表于 2023-11-29 02:56
\[a_n=\{3,6,3,1,2,1,2,1,2,1....\}\]
\

硬凑一个通项公式还是可以凑出来的

作者: 天山草 时间: 2023-11-29 18:24
这题为什么要这样叙述啊，本来很简单的题，却让人费解。

作者: 王守恩 时间: 2023-11-29 18:44
\(\frac{3+2=5}{5+1=6}\frac{6+2=8=4}{5+1=6=3}\frac{3+2=5=1}{4+1=5=1}\frac{1+2=3}{1+1=2}\frac{2+2=4=1}{3+1=4=1}\frac{1+2=3}{1+1=2}\frac{2+2=4=1}{3+1=4=1}\frac{1+2=3}{1+1=2}\frac{2+2=4=1}{3+1=4=1}\cdots\cdots\)

作者: 王守恩 时间: 2023-12-1 09:17
\(类似的题目, 需要的是耐心。\)
\(a_{1}=\frac{\sqrt{3}}{2},b_{1}=\frac{1}{2},\)
\(a_{n+1}=\frac{\sqrt{3}}{2}a_{n}-\frac{1}{2}b_{n}\)
\(b_{n+1}=\frac{1}{2}a_{n}+\frac{\sqrt{3}}{2}b_{n}\)
\(求: a_{1999},b_{1999}\)

作者: 王守恩 时间: 2023-12-2 07:53

Treenewbee 发表于 2023-11-28 03:57
复杂了，可简化为

\(12次才出现循环。这是13楼的题目,  6楼的公式还可以用吗？\)

\(\ a_{01}=\frac{\sqrt{3}}{2},b_{01}=\frac{1}{2},\)
\(\ a_{02}=\frac{\sqrt{3}}{2}a_{01}-\frac{1}{2}b_{01}=\frac{1}{2},\ \  b_{02}=\frac{1}{2}a_{01}+\frac{\sqrt{3}}{2}b_{01}=\frac{\sqrt{3}}{2}\)
\(\ a_{03}=\frac{\sqrt{3}}{2}a_{02}-\frac{1}{2}b_{02}=\frac{0}{2},          \ \ b_{03}=\frac{1}{2}a_{02}+\frac{\sqrt{3}}{2}b_{02}=\frac{2}{2}\)
\(\ a_{04}=\frac{\sqrt{3}}{2}a_{03}-\frac{1}{2}b_{03}=\frac{-1}{2},       b_{04}=\frac{1}{2}a_{03}+\frac{\sqrt{3}}{2}b_{03}=\frac{\sqrt{3}}{2}\)
\(\ a_{05}=\frac{\sqrt{3}}{2}a_{04}-\frac{1}{2}b_{04}=\frac{\sqrt{3}}{-2},b_{05}=\frac{1}{2}a_{04}+\frac{\sqrt{3}}{2}b_{04}=\frac{1}{2}\)
\(\ a_{06}=\frac{\sqrt{3}}{2}a_{05}-\frac{1}{2}b_{05}=\frac{-2}{2},       b_{06}=\frac{1}{2}a_{05}+\frac{\sqrt{3}}{2}b_{05}=\frac{0}{2}\)
\(\ a_{07}=\frac{\sqrt{3}}{2}a_{06}-\frac{1}{2}b_{06}=\frac{\sqrt{3}}{-2},b_{07}=\frac{1}{2}a_{06}+\frac{\sqrt{3}}{2}b_{06}=\frac{-1}{2}\)
\(\ a_{08}=\frac{\sqrt{3}}{2}a_{07}-\frac{1}{2}b_{07}=\frac{-1}{2},       b_{08}=\frac{1}{2}a_{07}+\frac{\sqrt{3}}{2}b_{07}=\frac{\sqrt{3}}{-2}\)
\(\ a_{09}=\frac{\sqrt{3}}{2}a_{08}-\frac{1}{2}b_{08}=\frac{0}{2},       \ \ b_{09}=\frac{1}{2}a_{08}+\frac{\sqrt{3}}{2}b_{08}=\frac{-2}{2}\)
\(\ a_{10}=\frac{\sqrt{3}}{2}a_{09}-\frac{1}{2}b_{09}=\frac{1}{2},    \ \    b_{10}=\frac{1}{2}a_{09}+\frac{\sqrt{3}}{2}b_{09}=\frac{\sqrt{3}}{-2}\)
\(\ a_{11}=\frac{\sqrt{3}}{2}a_{10}-\frac{1}{2}b_{10}=\frac{\sqrt{3}}{2},b_{11}=\frac{1}{2}a_{10}+\frac{\sqrt{3}}{2}b_{10}=\frac{-1}{2}\)
\(\ a_{12}=\frac{\sqrt{3}}{2}a_{11}-\frac{1}{2}b_{11}=\frac{2}{2},    \ \ b_{12}=\frac{1}{2}a_{11}+\frac{\sqrt{3}}{2}b_{11}=\frac{0}{2}\)
\(\ a_{13}=\frac{\sqrt{3}}{2}a_{12}-\frac{1}{2}b_{12}=\frac{\sqrt{3}}{2},b_{13}=\frac{1}{2}a_{12}+\frac{\sqrt{3}}{2}b_{12}=\frac{1}{2}\)
\(\ a_{14}=\frac{\sqrt{3}}{2}a_{13}-\frac{1}{2}b_{13}=\frac{1}{2},    \ \    b_{14}=\frac{1}{2}a_{13}+\frac{\sqrt{3}}{2}b_{13}=\frac{\sqrt{3}}{2}\)

作者: Treenewbee 时间: 2023-12-2 09:20

王守恩发表于 2023-12-1 01:17
\(类似的题目, 需要的是耐心。\)
\(a_{1}=\frac{\sqrt{3}}{2},b_{1}=\frac{1}{2},\)
\(a_{n+1}=\frac{\sq ...

显然\[a_{n+1}^2+b_{n+1}^2=a_n^2+b_n^2=...=a_1^2+b_1^2=1\]
\[ 令a_n=cos \theta_n,b_n=sin \theta_n\]
\[ \theta_1=\frac{\pi}{6}\]
\[ cos \theta_{n+1}+isin \theta_{n+1}=cos\frac{\pi}{6}cos \theta_{n}-sin\frac{\pi}{6}sin\theta_n+i(sin\frac{\pi}{6}cos \theta_{n}+cos\frac{\pi}{6}sin\theta_n)=(cos\frac{\pi}{6}+isin\frac{\pi}{6})(cos\theta_n+isin\theta_n)=cos(\frac{\pi}{6}+\theta_n)+isin(\frac{\pi}{6}+\theta_n)\]
\[\theta_{n+1}=\frac{\pi}{6}+\theta_n\]
\[\theta_n=\frac{n\pi}{6}\]
\[a_{1999}=cos\frac{1999\pi}{6}=cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2},b_{1999}=\sin \frac{1999 \pi }{6}=\sin \frac{7\pi }{6}=-\frac12\]

作者: Treenewbee 时间: 2023-12-2 09:24

王守恩发表于 2023-12-1 23:53
\(12次才出现循环。这是13楼的题目, 6楼的公式还可以用吗？\)

\(\ a_{01}=\frac{\sqrt{3}}{2},b_{01} ...

Last@NestList[{Sqrt[3]/2#[[1]]-1/2#[[2]],1/2#[[1]]+Sqrt[3]/2#[[2]]}&,{Sqrt[3]/2,1/2},1999-1]

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\[\left\{-\frac{\sqrt{3}}{2},-\frac{1}{2}\right\}\]

作者: 王守恩 时间: 2023-12-2 10:24

Treenewbee 发表于 2023-12-2 01:20
显然\[a_{n+1}^2+b_{n+1}^2=a_n^2+b_n^2=...=a_1^2+b_1^2=1\]
\[ 令a_n=cos \theta_n,b_n=sin \theta_n ...

观察数据，易得，每迭代一次，相当于把（an，bn）这个向量逆时针旋转了30度。

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