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

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

∵數列{a[n]}的通項a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n項和為S[n]∴a[n]=n^2Cos(2nπ/3)∴S[n]=1^2(-1/2)+2^2(-1/2)+3^2+4^2(-1/2)+5^2(-1/2)+6^2+…+n^2Cos(2nπ/3)當n=3k-2,即:k=(n+2)/3時:S[3k-2]=(-1/2…