问题
填空题
等差数列{an}的前n项和为Sn，a3=20，S3=36，则
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答案
∵a3=a1+2d=20，S3=3a1+3d=36
∴d=8，a1=4
∴Sn=4n+
×8=4n2n(n-1) 2
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- 1 2
+1 3
-1 3
+…+1 5
-1 29
)=1 31 15 31
故答案为：15 31