一道高一数学关于数列题目

∵a[n]=n^2*cos(2nπ/3)
∴S[n]
=(-1/2)*1^2+(-1/2)*2^2+3^2+(-1/2)*4^2+(-1/2)*5^2+6^2+...+n^2*cos(2nπ/3)
=(-1/2)(1^2+2^2+3^2+...+n^2)+(3/2)3^2(1^2+2^2+...+INT(n/3)^2)
【INT为取整函数】
∵1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
∴S[n]
=-n(n+1)(2n+1)/12+(3/2)3^2*INT(n/3)[INT(n/3)+1][2INT(n/3)+1]/6
=-n(n+1)(2n+1)/12+3^2*INT(n/3)[INT(n/3)+1][2INT(n/3)+1]/4