(Ⅰ)若q=1,则5S 2=10a 1,4S 4=16a 1,∵a 1≠0,
∴5S 2≠4S 4,不合题意.(2分)
若q≠1,由5S 2=4S 4得 5×
a 1 (1- q 2 )
1-q =4×
a 1 (1- q 4 )
1-q ,
∴ q 2 =
1
4 ,又q>0,
∴ q=
1
2 ..(5分)
(Ⅱ) b n =
1
2 +
a 1 [1- (
1
2 ) n-1 ]
1-
1
2 =
1
2 +2 a 1 - a 1 •(
1
2 ) n-2 ,(7分)
由b n为等比数列知:
1
2 +2 a 1 =0 ,得 a 1 =-
1
4 ,
∴ b n =
1
4 •(
1
2 ) n-2 =
1
2 n .(9分)
则 T n =
1
2 +
3
2 2 +
5
2 3 +…+
2n-1
2 n ①
1
2 T n =
1
2 2 +
3
2 2 +…+
2n-3
2 n +
2n-1
2 n+1 ②
两式相减化简得T n=3-
2n+3
2 n (12分)