又一條中二既數學問題!

2007-09-28 3:54 am

3^(2n-1)....+4(3^2n)除2(3^2n+1)=?
3^(2n-1)..再加4(3^2n)...除以2(3^2n+1)..

2^(2n+1)..+7(2^n)除4(2^n)=?

9(4^n)....-3(2^n+1)除5(2^n-1)=?

回答 (2)

NCY

2007-09-28 4:12 am

✔ 最佳答案

1) 32n-1)+4(32n)/2(32n+1)
= 32n‧3^(-1) + 4(3^2n) / 2(3)(32n)
= 32n(1/3+4) / (32n)(6)
= (13/3) / 6
= 13/18

2. [22n+1 +7(2n)]/4(2n)
= 2n‧2n‧2 + 7(2n) / 4(2n)
= 2n(2‧2n + 7)/4(2n)
= (2‧2n + 7) / 4

3.[9(4n) - 3(2n+1)] / 5(2n-1)
= [9(22)n - 3(21)(2n)] / 5(2n/21) ←應用指數定律 am+n = aman
= (21)[9(22)n - 3(2)(2n)] / 5(2n) ←應用指數定律 am-n = am / an
= 2[9(22n) - 6(2n)] / 5(2n) ←應用指數定律 (am)n = (an)m = amn
= 6[3(2n+n) - 2(2n)] / 5(2n)
= 6[3(2n)(2n) - 2(2n)] / 5(2n)
= 6(2n)[3(2n) - 2] / 5(2n) ←約去 2n
=6[3(2n) - 2]/5
= [18(2n) - 12] / 5

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用戶128885

2007-09-30 8:41 am

好詳細,又好易明白

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