數列an的通項an=n²;(cos²;nπ/3-sin²;nπ/3),其前n項的和為sn,則S30=?

數列an的通項an=n²;(cos²;nπ/3-sin²;nπ/3),其前n項的和為sn,則S30=?

an=n²;(cos²;nπ/3-sin²;nπ/3)=n^2*cos(2nπ/3)(二倍角公式)cos(2π/3)=-1/2cos(4π/3)=-1/2cos(6π/3)=1所以a(3k-2)+a(3k-1)+a(3k)=(3k-2)^2*(-1/2)+(3k-1)^2*(-1/2)+(3k)^2*1= 9k-5/2所以S30=a1+…