已知数列{an}对于任意的p、q∈N*,满足ap+q=ap+aq且a2=2,则1a1a2+1a2a3+…+1a2008a2009=______.-数学

题目简介

已知数列{an}对于任意的p、q∈N*,满足ap+q=ap+aq且a2=2,则1a1a2+1a2a3+…+1a2008a2009=______.-数学

题目详情

已知数列{an}对于任意的p、q∈N*,满足ap+q=ap+aq且a2=2,则
1
a1a2
+
1
a2a3
+…+
1
a2008a2009
=______.
题型:填空题难度:中档来源:不详

答案

数列{an}对于任意的p、q∈N*,满足ap+q=ap+aq且a2=2,所以a2=a1+a1且a1=1,
所以an+1=an+1,数列是等差数列,an=n,所以class="stub"1
a1a2
+class="stub"1
a2a3
+…+class="stub"1
a2008a2009

=-(class="stub"1
a2
- class="stub"1
a1
+class="stub"1
a3
-class="stub"1
a2
+…+class="stub"1
a2009
-class="stub"1
a2008
)=1-class="stub"1
2009
=class="stub"2008
2009

故答案为:class="stub"2008
2009

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