问题 选择题
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
答案

∵数列{an}满足a1=a2=1,an+an+1+an+2=cos

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

=cos

2nπ
3

=cos(2nπ-

3

=cos(-

3

=cos

3

=-cos

π
3

=-

1
2

∵2013÷3=671,即S2013正好是前671组的和,

∴S2013=-

1
2
×671=-
671
2

故选D.

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