已知数列{an},满足a1=1,1an+1=1an+1,Sn是数列{anan+1}的前n项和,则S2011=______.-数学

题目简介

已知数列{an},满足a1=1,1an+1=1an+1,Sn是数列{anan+1}的前n项和,则S2011=______.-数学

题目详情

已知数列{an},满足a1=1,
1
an+1
=
1
an
+1
,Sn是数列{anan+1}的前n项和,则S2011=______.
题型:填空题难度:偏易来源:不详

答案

a1=1,class="stub"1
an+1
=class="stub"1
an
+1

{class="stub"1
an
}
是以1为首项以1为公差的等差数列
根据等差数列的通项公式可得,class="stub"1
an
=n
an=class="stub"1
n

anan+1=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

S2011=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
2011
-class="stub"1
2012

=1-class="stub"1
2012
=class="stub"2011
2012

故答案为:class="stub"2011
2012

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