Question

等差数列{a_n}的前n项和为S_n，a₃=20，S₃=36，则

1

S1-1

+

1

S2-1

+

1

S3-1

+…+

1

S15-1

=______．

Answer 1

∵a₃=a₁+2d=20，S₃=3a₁+3d=36
∴d=8，a₁=4
∴Sn=4n+

n(n-1)

2

×8=4n2
而

1

Sn-1

=

1

4n2- 1

=

1

(2n+1)(2n-1)

=

1

2

(

1

2n-1

-

1

2n+1

)
则

1

S1-1

+

1

S2-1

+

1

S3-1

+…+

1

S15-1

=

1

1×3

+

1

3×5

+…+

1

29×31

=

1

2

( 1-  

1

3

+

1

3

-

1

5

+…+

1

29

-

1

31

)=

15

31

故答案为：

15

31