题目

数列{an}的通项公式为an=2nsin(nπ/2-π/3)+√3ncos(nπ/2),前n项和为Sn,
求S2013

提问时间：2021-03-12

答案

an=2nsin(nπ/2-π/3)+√3ncos(nπ/2)
= 2n [ sin(nπ/2)cos(π/3) - cos(nπ/2)sin(π/3) ] +√3ncos(nπ/2)
=nsin(nπ/2)
ie
an = n ; n=1,5,9,...
=0 ; n=2,4,6,...
= -n ; n=3,7,11,...
S2013
=a1+a2+...+a2013
=(a1+a5+a9+...+a2013)+(a2+a4+...+a2012)+(a3+a7+...+a2011)
=(1+5+...+2013)-(3+7+...+2011)
=(2013+1)504/2-(2011+3)503/2
=1007