(1)∵2S n=
1
2 (a n 2+a n),2S n+1=
1
2 (a n+1 2+a n+1)
∴两式相减可得(a n+1+a n)(a n+1-a n-1)=0,
∵数列{a n}各项均正,
∴a n+1-a n=1,
∴{a n}是以1为公差的等差数列,
∵2S 1=
1
2 (a 1 2+a 1),
∴a 1=1
∴a n=n;
(2)b n=
1
2 (
1
n -
1
n+2 )
∴T n=
1
2 (1-
1
3 +
1
2 -
1
4 +…+
1
n -
1
n+2 ) =
1
2 ( 1+
1
2 -
1
n+1 -
1
n+2 )=
n(3n+5)
2(n+1)(n+2) .