数列{an}满足a1=32,an+1=an2-an+1(n∈N*),则m=1a1+1a2+1a3+…+1a2009的整数部分是()A.3B.2C.1D.0-数学

题目简介

数列{an}满足a1=32,an+1=an2-an+1(n∈N*),则m=1a1+1a2+1a3+…+1a2009的整数部分是()A.3B.2C.1D.0-数学

题目详情

数列{an}满足a1=
3
2
,an+1=an2-an+1(n∈N*),则m=
1
a1
+
1
a2
+
1
a3
+…+
1
a2009
的整数部分是(  )
A.3B.2C.1D.0
题型:单选题难度:偏易来源:不详

答案

由题设知,an+1-1=an(an-1),
class="stub"1
an+1-1
=class="stub"1
an(an-1)
=class="stub"1
an-1
-class="stub"1
an

class="stub"1
an-1
-class="stub"1
an+1-1
=class="stub"1
an

通过累加,得
m=class="stub"1
a1
+class="stub"1
a2
+class="stub"1
a3
+…+class="stub"1
a2009
=class="stub"1
a1-1
-class="stub"1
a2010-1
=2-class="stub"1
a2010-1

由an+1-an=(an-1)2≥0,
即an+1≥an,
a1=class="stub"3
2

a2=class="stub"7
4

得a3=class="stub"37
16

∴a2010≥a2009≥a2008≥a3>2,
0<class="stub"1
a2010-1
<1

∴1<m<2,
所以m的整数部分为1.
故选C.

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