(1)由 S n + S n-2 =2 S n-1 + 2 n-1 得 a n = a n-1 + 2 n-1 (n≥3,n∈N*) ,
∵a 2=5,∴当n≥3时,a n=a 2+(a 3-a 2)+(a 4-a 3)+…+(a n-a n-1)=5+2 2+2 3+…+2 n-1=2 n+1,
经验证a 1=3,a 2=5也符合上式,
∴ a n = 2 n +1(n∈ N * ) ;
(2)由(1)可得 b n =
n
a n -1 =
n
2 n ,
∴ T n =
1
2 +
2
2 2 +
3
2 3 +…+
n
2 n ① ⇒
1
2 T n =
1
2 2 +
2
2 3 +…+
n-1
2 n +
n
2 n+1 ②,
①-②有:
1
2 T n =
1
2 +
1
2 2 +
1
2 3 +…+
1
2 n -
n
2 n+1 =1-
1
2 n -
n
2 n+1 ,
∴ T n =2-
n+2
2 n ;
(3)∵ f(x)= 2 x-1 , c n =
1
a n a n+1 ,
∴ c n f(n)=
2 n-1
( 2 n +1)( 2 n+1 +1) =
1
2 (
1
2 n +1 -
1
2 n+1 +1 )(n∈N*) ,
∴Q n=c 1f(1)+c 2f(2)+…+c nf(n)
=
1
2 [(
1
2 1 +1 -
1
2 2 +1 )+(
1
2 2 +1 -
1
2 3 +1 )+…+(
1
2 n +1 -
1
2 n+1 +1 )]
=
1
2 (
1
1+2 -
1
2 n+1 +1 )<
1
2 ×
1
3 =
1
6 .