数列{an}满足a1=a2=1,an+an+1+an+2=cos2nπ/3若数列{an}的前n项和为sn则s2012的值

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数列{an}满足a1=a2=1,an+an+1+an+2=cos2nπ/3若数列{an}的前n项和为sn则s2012的值
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数列{an}满足a1=a2=1,an+an+1+an+2=cos2nπ/3若数列{an}的前n项和为sn则s2012的值
数列{an}满足a1=a2=1,an+an+1+an+2=cos2nπ/3若数列{an}的前n项和为sn则s2012的值

数列{an}满足a1=a2=1,an+an+1+an+2=cos2nπ/3若数列{an}的前n项和为sn则s2012的值
从第一项开始,3个一组,则第n组的第一个数为a(3n-2)
a(3n-2)+a(3n-1)+a(3n)
=cos[2(3n-2)π/3]
=cos(2nπ -4π/3)
=cos(-4π/3)
=cos(4π/3)
=-cos(π/3)
=-1/2
2012÷3=670.2,即S2012=670*(-1/2)+S2
S2012=-335+ 1+1=-333
提示:本题关键是找到规律,3个一组分组后,每组的和都是定值-1/2,剩下的就比较简单了.

答案是672,求解