（1）设等差数列{a_n}的公差为d，∵S₅=45，S₆=60，∴

5a1+ 5×4 2 d=45 6a1+ 6×5 2 d=60

，解得

a1=5 d=2

．∴a_n=5+（n-1）×2=2n+3．

（2）∵b_n+1-b_n=a_n=2n+1，b₁=3，

∴b_n=（b_n-b_n-1）+（b_n-1-b_n-2）+…+（b₂-b₁）+b₁

=[2（n-1）+3]+[2（n-2）+3]+…+（2×1+3）+3

=2×

n(n-1)

2

+3n

=n²+2n．

∴

1

bn

=

1

n(n+2)

=

1

2

(

1

n

-

1

n+2

)．

∴T_n=

1

2

[(1-

1

3

)+(

1

2

-

1

4

)+(

1

3

-

1

5

)+…+(

1

n-1

-

1

n+1

)+(

1

n

-

1

n+2

)]

=

1

2

(1+

1

2

-

1

n+1

-

1

n+2

)

=

3

4

-

1

2(n+1)

-

1

2(n+2)

．