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
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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
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
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
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
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 ...
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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
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