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)

天山草

这题为什么要这样叙述啊，本来很简单的题，却让人费解。

王守恩

\(\frac{3+2=5}{5+1=6}\frac{6+2=8=4}{5+1=6=3}\frac{3+2=5=1}{4+1=5=1}\frac{1+2=3}{1+1=2}\frac{2+2=4=1}{3+1=4=1}\frac{1+2=3}{1+1=2}\frac{2+2=4=1}{3+1=4=1}\frac{1+2=3}{1+1=2}\frac{2+2=4=1}{3+1=4=1}\cdots\cdots\)

王守恩

\(类似的题目, 需要的是耐心。\)
\(a_{1}=\frac{\sqrt{3}}{2},b_{1}=\frac{1}{2},\)
\(a_{n+1}=\frac{\sqrt{3}}{2}a_{n}-\frac{1}{2}b_{n}\)
\(b_{n+1}=\frac{1}{2}a_{n}+\frac{\sqrt{3}}{2}b_{n}\)
\(求: a_{1999},b_{1999}\)

王守恩

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

\(12次才出现循环。这是13楼的题目,  6楼的公式还可以用吗？\)

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

Treenewbee

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

Treenewbee

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} ...

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

复制代码

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

王守恩

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。