(I)∵函数f(x)=a•b x的图象过点
A(0,
1
16 ),B(2,
1
4 )
∴
a• b 0 =
1
16
a• b 2 =
1
4 解得:a=
1
16 ,b=2,∴f(x)=2 x-4
(II)a n=log 2 f(n)=
log 2 n-4 2 =n-4
∴{a n}是首项为-3,公差为1的等差数列
∴S n=-3n+
1
2 n(n-1)=
1
2 n(n-7);
(III)b n=a n(
1
2 ) n =(n-4) (
1
2 ) n
T n=-3×
1
2 +(-2)× (
1
2 ) 2 +…+(n-4)× (
1
2 ) n ①
1
2 T n =-3× (
1
2 ) 2 +(-2)× (
1
2 ) 3 +…+(n-4)× (
1
2 ) n-1 ②
①-②,得:
1
2 T n=-3×
1
2 + (
1
2 ) 2 + (
1
2 ) 3 +…+ (
1
2 ) n -(n-4)× (
1
2 ) n-1
∴T n=-2-(n-2) (
1
2 ) n .