急!!!!指數定律

1)
9(4^n)-3(2^(n+1))/5(2^(n-1))
=9(2^2n) - 3(2^(n+1))/5(2^(n-1)) '.'(2^(n+1)/2^(n-1) = 4)
=9(2^2n) - 3*4/5
= 9(2^2n) -12 /5

2007-09-06 22:56:14 補充：
9(2^2n) -12 /5 = 18(2^n)-12/5 上面位仁兄答案錯左因為佢當左9(4^n)-3(2^(n 1))/5(2^(n-1)係 〔9(4^n)-3(2^(n 1))〕/5(2^(n-1)

2007-09-06 22:57:16 補充：
上面位仁兄其實9(4^n)-3(2^(n 1))/5(2^(n-1)　係　9(4^n)　-　〔3(2^(n 1))/5(2^(n-1)〕

9(4^n)-3(2^(n+1))/5(2^(n-1)
= 9(2^2n) - 3(2^n)(2) / 5(2^n)(2^(-1))
=9(2^n)(2^n)- 6(2^n) / (5/2)(2^n)
= 2^n(9．2^n - 6) / (5/2)(2^n)
=(9．2^n - 6) / (5/2)
=2(9．2^n - 6)/5
= (18．2^n - 12)/5