(1)∵点(n,S n)(n∈N *)均在函数y=f(x)的图象上,
∴ S n =
1
2 n 2 +
3
2 n ,
∴当n=1时, a 1 = S 1 =
1
2 × 1 2 +
3
2 ×1=2 ;
当n≥2时,a n=S n-S n-1=
1
2 n 2 +
3
2 n- [
1
2 (n-1 ) 2 +
3
2 (n-1)]=n+1 .
当n=1时,也适合上式,
因此 a n =n+1(n∈ N * ) .
(2)由(1)可得: b n =
a n
2 n-1 =
n+1
2 n-1 .
∴T n=
2
2 0 +
3
2 1 +
4
2 2 +…+
n
2 n-2 +
n+1
2 n-1 ,
1
2 T n =1+
3
2 2 +…+
n
2 n-1 +
n+1
2 n ,
两式相减得
1
2 T n =2+
1
2 +
1
2 2 +…+
1
2 n-1 -
n+1
2 n =1+
1×(1-
1
2 n )
1-
1
2 -
n+1
2 n =3 -
1
2 n-1 -
n+1
2 n
∴ T n =6-
n+3
2 n-1 .
(3)证明:由c n=
a n
a n+1 +
a n+1
a n =
n+1
n+2 +
n+2
n+1 >2
n+1
n+2 •
n+2
n+1 =2,
∴c 1+c 2+…+c n>2n.
又c n=
n+1
n+2 +
n+2
n+1 =2+
1
n+1 -
1
n+2 ,
∴c 1+c 2+…+c n=2n+[(
1
2 -
1
3 )+(
1
3 -
1
4 )+…+(
1
n+1 -
1
n+2 )]=2n+
1
2 -
1
n+2 <2n+
1
2 .
∴2n<c 1+c 2+…+c n<2n+
1
2 成立.