数列{an}满足a1=1,且对任意的正整数m,n都有am+n=am+an+mn,则1a1+1a2+…+1a2012+1a2013=______.-数学

题目简介

数列{an}满足a1=1,且对任意的正整数m,n都有am+n=am+an+mn,则1a1+1a2+…+1a2012+1a2013=______.-数学

题目详情

数列{an}满足a1=1,且对任意的正整数m,n都有am+n=am+an+mn,则
1
a1
+
1
a2
+…+
1
a2012
+
1
a2013
=______.
题型:填空题难度:偏易来源:不详

答案

令n=1,得an+1=a1+an+n=1+an+n,∴an+1-an=n+1
用叠加法:an=a1+(a2-a1)+…+(an-an-1)=1+2+…+n=
n(n+1)
2

所以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
a2012
+class="stub"1
a2013
=2(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
2013
-class="stub"1
2014
)
=2×class="stub"2013
2014
=class="stub"2013
1007

故答案为:class="stub"2013
1007

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