数列{an}满足a1=a2=1,an+an+1+an+2=cos2nπ3(n∈N*),若数列{an}的前n项和为Sn,则S2013的值为()A.2013B.671C.-671D.-6712-数学

题目简介

数列{an}满足a1=a2=1,an+an+1+an+2=cos2nπ3(n∈N*),若数列{an}的前n项和为Sn,则S2013的值为()A.2013B.671C.-671D.-6712-数学

题目详情

数列{an}满足a1=a2=1,an+an+1+an+2=cos
2nπ
3
(n∈N*)
,若数列{an}的前n项和为Sn,则S2013的值为(  )
A.2013B.671C.-671D.-
671
2
题型:单选题难度:偏易来源:闵行区一模

答案

∵数列{an}满足a1=a2=1,an+an+1+an+2=cosclass="stub"2nπ
3
(n∈N*)

∴从第一项开始,3个一组,则第n组的第一个数为a3n-2
a3n-2+a3n-1+a3n
=cosclass="stub"2nπ
3

=cos(2nπ-class="stub"4π
3

=cos(-class="stub"4π
3

=cosclass="stub"4π
3

=-cosclass="stub"π
3

=-class="stub"1
2

∵2013÷3=671,即S2013正好是前671组的和,
∴S2013=-class="stub"1
2
×671=-class="stub"671
2

故选D.

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