数列{an}满足:a1=1,且对任意的m,n∈N*都有:am+n=am+an+mn,则1a1+1a2+1a3+…+1a2011=()A.20102011B.20111006C.20112012D.20

题目简介

数列{an}满足:a1=1,且对任意的m,n∈N*都有:am+n=am+an+mn,则1a1+1a2+1a3+…+1a2011=()A.20102011B.20111006C.20112012D.20

题目详情

数列{an}满足:a1=1,且对任意的m,n∈N*都有:am+n=am+an+mn,则
1
a1
+
1
a2
+
1
a3
+…+
1
a2011
=(  )
A.
2010
2011
B.
2011
1006
C.
2011
2012
D.
2010
1006
题型:单选题难度:中档来源:不详

答案

因为an+m=am+an+mn对任意的m,n∈N*都成立
所以an+1=an+a1+n=1+n
即an+1-an=1+n
所以a2-a1=2
   a3-a2=3

   an-an-1=n
把上面n-1个式子相加可得,an-a1=2+3+4+…+n
所以an=1+2+3+…+n=
n(n+1)
2

从而有class="stub"1
an
=class="stub"2
n(n+1)
=2(class="stub"1
n
-class="stub"1
n+1
)

所以class="stub"1
a1
+class="stub"1
a2
+…+class="stub"1
an
=2(1-class="stub"1
2
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n
-class="stub"1
n+1
)
=2(1-class="stub"1
n+1
)=class="stub"2n
n+1

class="stub"1
a1
+class="stub"1
a2
+…+class="stub"1
a2011
=class="stub"2×2011
2012
=class="stub"2011
1006

故选:B

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