已知数列{an},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=()A.n(n+1)2B.2n(n+1)C.2nn+1D.

题目简介

已知数列{an},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=()A.n(n+1)2B.2n(n+1)C.2nn+1D.

题目详情

已知数列{an},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
=(  )
A.
n(n+1)
2
B.
2
n(n+1)
C.
2n
n+1
D.
n
2(n+1)
题型:单选题难度:中档来源:不详

答案

∵点P(an,an+1)(n∈N*)在直线x-y+1=0上
∴an-an+1+1=0
∴数列{an}是以1为首项,以1为公差的等差数列.
∴an=n
sn=
n(n+1)
2

class="stub"1
sn
=class="stub"2
n(n+1)
=2(class="stub"1
n
-class="stub"1
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
3
+…+class="stub"1
n
-class="stub"1
n+1
)
=class="stub"2n
n+1

故选C

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