数列{an}的通项公式an=ncos
,其前n项和为Sn,则S2012等于( )
(A)1006 (B)2012 (C)503 (D)0
数列{an}的通项公式an=ncos,其前n项和为Sn,则S2012等于( )
(A)1006 (B)2012 (C)503 (D)0
A解析:∵当n∈N*时,
a4k+1=(4k+1)cos(2kπ+
)=0,
a4k+2=(4k+2)cos(2kπ+π)=-(4k+2),
a4k+3=(4k+3)cos(2kπ+
)=0,
a4n+4=(4k+4)cos(2kπ+2π)=4k+4,
∴a4k+1+a4k+2+a4k+3+a4k+4=2.
则S2012=(a1+a2+a3+a4)+(a5+a6+a7+a8)+…+(a2009+a2010+a2011+a2012)=2×503=1006.