Question

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

求解释，谢谢

Answer 1

解：
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。

Answer 2

答案是672，求解