数列{an}满足a1=1,an+1=an+n+1(n∈N*),则1a
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数列{an}满足a1=1,an+1=an+n+1(n∈N*),则
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∵{an}满足a1=1,an+1=an+n+1(n∈N*),
a2-a1=1+1, a3-a2=2+1, a4-a3=3+1, … an-an-1=(n-1)+1, ∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1) =1+(1+1)+(2+1)+(3+1)+…+[(n-1)+1] =n+1+2+3+…+(n-1) = n(n+1) 2, ∴ 1 an= 2 n(n+1)=2( 1 n− 1 n+1), ∴ 1 a1+ 1 a2+…+ 1 a2013 =2(1- 1 2+ 1 2- 1 3+…+ 1 2013− 1 2014) =2(1- 1 2014) = 2013 1007. 故答案为: 2013 1007.
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