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
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
∵2010=670×3
∴a2010+a2011+a2012=cos(2010×2π/3)=cos(670×2π)=cos(2π)=1
同理:
a2007+a2008+a2009=cos(2007×2π/3)=cos(2π)=1
.
a3+a4+a5=cos(2π)=1
∴s2012=(a2012+a2011+a2010)+(a2009+a2008+a2007)+.+(a5+a4+a3)+a2+a1
=670×1+a2+a1
=670+2
=672
从第一项开始 3个一组分组 每组的和都是定值-1/2 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
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