证明：（1）∵

an

an+1

=

4an+2

an+1+2

∴2a_n-2a_n+1=3a_na_n+1

两边同时除以a_na_n+1可得，

1

an+1

-

1

an

=

3

2

∴数列列{

1

an

}是以

1

a1

=

5

2

为首项，以

3

2

为公差的等差数列，

∴

1

an

=

5

2

+

3

2

(n-1)=

3n+2

2

∴a_n=

2

3n+2

（2）bn=an•an+1=

2

3n+2

•

2

3n+5

=

4

3

(

1

3n+2

-

1

3n+5

)

∴Tn=b1+b2+b3+…+bn=

4

3

(

1

5

-

1

3n+5

)＜

4

15