已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则1S1+1S2+1S3+…+1Sn=______.-数学

题目简介

已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则1S1+1S2+1S3+…+1Sn=______.-数学

题目详情

已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则
1
S1
+
1
S2
+
1
S3
+…+
1
Sn
=______.
题型:填空题难度:中档来源:不详

答案

∵点P(an,an+1)(n∈N*)在直线x-y+1=0上,
∴an+1-an=1,
∴数列{an}是等差数列,
∵a1=1,
∴sn=
n2+n
2

class="stub"1
sn
=class="stub"2
n(n+1)

class="stub"1
S1
+class="stub"1
S2
+class="stub"1
S3
+…+class="stub"1
Sn
=2(1-class="stub"1
2
+class="stub"1
2
-…-class="stub"1
n+1
)=class="stub"2n
n+1

故答案为class="stub"2n
n+1

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