数列an的通项an=n²(cos²nπ/3-sin²nπ/3),其前n项的和为sn,则Sn=?
问题描述:
数列an的通项an=n²(cos²nπ/3-sin²nπ/3),其前n项的和为sn,则Sn=?
答
∵数列{a[n]}的通项a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n项和为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)当n=3k-2,即:k=(n+2)/3时:S[3k-2]=(-1/2...