1.將長度為1的線段,剪去其長度的2/3;再將剩下的線段:剪去其長度的2/3;如此繼續下去,設an為剪了n次後剩下的長度.
1.寫出數列〈an〉的遞迴關係式
2.求a5=??
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高中數學數列與級數問題
2013-05-28 5:33 am
回答 (3)
2013-05-28 5:38 am
✔ 最佳答案
長度為1的線段 -->a1=1剪去其長度的2/3
剩1/3 -->a2=1/3
剩1/3
再剪去其長度的2/3-->只要1/3
(1/3)(1/3) -->a3=(1/3)(1/3)
依此類推
r=1/3
a1=1
a2=(1/3)(1/3)=(1/3)^2 ...^表次方
...
an= (1/3)^n
所以a5=(1/3)^5
2013-05-28 8:07 am
a_0 = 1
a_{n+1} = a_n*(1-2/3) = (1/3)a_n, n=0,1,2,3,...
a_5 = (1/3)a_4 = (1/3)^2 a_3 = ... = (1/3)^5 a_5 = (1/3)^5
a_{n+1} = a_n*(1-2/3) = (1/3)a_n, n=0,1,2,3,...
a_5 = (1/3)a_4 = (1/3)^2 a_3 = ... = (1/3)^5 a_5 = (1/3)^5
參考: 自己
2013-05-28 6:23 am
a_0 = 1
a_{n+1} = a_n*(1-2/3) = (1/3)a_n, n=0,1,2,3,...
a_5 = (1/3)a_4 = (1/3)^2 a_3 = ... = (1/3)^5 a_5 = (1/3)^5
a_{n+1} = a_n*(1-2/3) = (1/3)a_n, n=0,1,2,3,...
a_5 = (1/3)a_4 = (1/3)^2 a_3 = ... = (1/3)^5 a_5 = (1/3)^5
收錄日期: 2021-05-04 01:54:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130527000016KK04400