设数列b_n的前n项和为S_n

由题意可得an=

1+2+3+…+n

n

=

n(n+1) 2

n

=

n+1

2

∴an+1=

n+2

2

∴bn=

1

anan+1

=

1

n+1 2 n+2 2

=

4

(n+1)(n+2)

=4(

1

n+1

-

1

n+2

)

∴S_n=b₁+b₂+…+b_n-1+b_n

=4(

1

2

-

1

3

+ 

1

3

-

1

4

…+

1

n

-

1

n+1

+

1

n+1

-

1

n+2

)

=4(

1

2

-

1

n+2

)

=

2n

n+2

∴bn=

1

anan+1

的前n项和为

2n

n+2

．