這一題,我想問點樣把2[3^(2n-1)]變成18[3^(2n-1)]
同埋可否比一個詳細既答案!!
1) 3^(2n-1)+4(3^2n)/2[3^(2n+1)]
2) [2^(2n+1)+7(2^n)]/4(2^n)
指數定律(form3)數學
2007-09-07 6:19 am
回答 (1)
2007-09-07 7:10 am
✔ 最佳答案
2[3^(2n+1)] = 2[3^(2n-1)*3^2] = 2*3^2[3^(2n-1)] = 18[3^(2n-1)]1)
3^(2n-1)+4(3^2n)/2[3^(2n+1)]
= 3^(2n-1)+4(3^2n)/2{3^[(2n)+1]} ('.' (3^2n)/[3^(2n+1)] = 3^(2n-2n-1) =1/3)
=3^(2n-1)+ 4/ 2*3
=3^(2n-1)+2/3
2)
[2^(2n+1)+7(2^n)]/4(2^n)
= [2^n(2^{n+1}) + 7(2^n)]/4(2^n) ('.' 2^(2n+1) = 2^(n+1)*(2^n) )
分母分子一齊除2^n
= [2^(n+1)+7]/4
收錄日期: 2021-04-13 13:21:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070906000051KK04461