数列{an}满足a1=1，且对于任意的n∈N*都有an+1=an+a1+n，则1a1+1a2+…+1a2017等于（）A.20162017B.40322017C.20172018D.40342018

Question

数列{a_n}满足a₁=1，且对于任意的n∈N^*都有a_n+1=a_n+a₁+n，则

(1)/(a1)+(1)/(a2)+…+(1)/(a2017)等于（　　）

A.

(2016)/(2017)

B.

(4032)/(2017)

C.

(2017)/(2018)

D.

(4034)/(2018)

Answer 1

∵数列{a_n}满足a₁=1，且对于任意的n∈N^*都有a_n+1=a_n+a₁+n，
∴a_n+1-a_n=1+n，
∴a_n=a₁+（a₂-a₁）+（a₃-a₂）+…+（a_n-a_n-1）
=1+2+3+…+n
=(n(n+1))/(2)．
∴(1)/(an)=(2)/(n(n+1))=2((1)/(n)-(1)/(n+1))．
∴(1)/(a1)+(1)/(a2)+…+(1)/(a2017)=2[(1-(1)/(2))+((1)/(2)-(1)/(3))+…+((1)/(2017)-(1)/(2018))]
=2×(1-(1)/(2018))
=(2017)/(1009)=(4034)/(2018)．
故选：D．