数列{an}的通项公式为an=2nsin(nπ/2-π/3)+√3ncos(nπ/2),前n项和为Sn,则S2012=答案是-1006
问题描述：
数列{an}的通项公式为an=2nsin(nπ/2-π/3)+√3ncos(nπ/2),前n项和为Sn,则S2012=
答案是-1006
答
和角公式得an=nsin(nπ/2),sin(nπ/2)的值在1，0，-1，0中周期出现，2012/4=503,S2012=1-3+5-7+9-11+...+2009-2011=503*（-2）=-1006
答
an=2nsin(nπ/2-π/3)+√3ncos(nπ/2)
=2n[ (1/2)sin(nπ/2)-(√3/2)cos(nπ/2)] +√3ncos(nπ/2)
= nsin(nπ/2)
an = n if n=1,5,9,...
= 0 if n=2,4,6,8,10,.
= -n if n=3,7,11,.
S2012 = a1+a2+...+a2012
= (1+5+9+...+2009)-(3+7+9+...+2011)
= (1+2009)503/2 -(3+2011)503/2
=-503(2)
=-1006