（1）∵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]=...这是原式 ...an=n(cosnπ/3²-sinnπ/3²).......你回答的是哪道题啊