（1）由a₁=S₁=

1

4

(a1-1)(a1+3)，及a_n＞0，得a₁=3

（2）由Sn=

1

4

(an-1)(an+3)

得Sn-1=

1

4

(an-1-1)(an-1+3)．∴当n≥2时，

an=

1

4

(

a

2n

-

a

2n-1

)+2(an-an-1)

∴2（a_n+a_n-1）=（a_n+a_n-1）（a_n-a_n-1）

∵a_n+a_n-1＞0∴a_n-a_n-1=2，

∴由（1）知，{a_n}是以3为首项，2为公差的等差数列，∴a_n=2n+1．

（3）由（2）知S_n=n（n+2）∴bn=

1

Sn

=

1

2

(

1

n

-

1

n+2

)，

T_n=b₁+b₂+…+b_n

=

1

2

(1-

1

3

+

1

2

-

1

4

++

1

n-1

-

1

n+1

+

1

n

-

1

n+2

)

=

1

2

[

3

2

-

2n+3

(n+1)(n+2)

]

=

3

4

-

2n+3

2(n+1)(n+2)