請詳細解釋下條計法 :
圖片參考：http://imgcld.yimg.com/8/n/HA05788109/o/20131126125204.jpg
答案計法 :T(2) =1 = (1/2)^0
T(4) = 1/2 = (1/2)^1
T(6) = 1/4 = (1/2)^2
T(8) = 1/8 = (1/2)^3
T(2n) = 1/2^(n-1) = (1/2)^(n-1) [即點解不是T(2n) = 1/2^(2n-1)]
T(2)*T(4)*...*T(2n)
= (1/2)^0 * (1/2)^1 * ... * (1/2)^(n-1)
= (1/2)^[0+1+...+(n-1)]
= (1/2)^[n(n-1)/2] [點解 (1/2)^[0+1+...+(n-1)] 變成 (1/2)^[n(n-1)/2]及多了分母2 ???]
= 2^[-n(n-1)/2]
急! 數學 sequences 1條
2013-11-26 8:54 pm
回答 (2)
2013-11-26 9:44 pm
✔ 最佳答案
first term is 2^(1/2)common ratio is 1 / 2^(1/2)
so T(2)= 2^(1/2) * (1 / 2^(1/2))^(2-1)=1 = (1 / 2^(1/2))^0
T(4)= 2^(1/2) * (1 / 2^(1/2))^(4-1)=1/2 = (1 / 2^(1/2))^2
.........
T(2n)= 2^(1/2) * (1 / 2^(1/2))^(2n-1)=(1 / 2^(1/2))^(2n-2)
T(2)*T(4)*....*T(2n)
= (1 / 2^(1/2))^(0+2+...+(2n-2))
= (1 / 2^(1/2))^((2n-2)n/2) (頭+尾)(項數)/2 (0+2n-2)(n)/2
= (1 / 2^(1/2))^(n(n-1))
= 2^(-(1/2)*n(n-1))
=2^(-n(n-1)/2)
2013-11-27 1:37 am
收錄日期: 2021-04-13 19:49:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131126000051KK00045