可设数列{bn}.bn=cos(2nπ/3),n=1,2,3,…则an=n²bn.
易知,bn=Cos(2nπ/3)的周期是3.
∴b1=b4=b7=…=b28=-1/2.
b2=b5=b8=,=b29=-1/2.
b3=b6=b9=,=b30=1=(-1/2)+(3/2).
∴S30=∑an=∑n²bn=(-1/2)(1²+2²+3²+…+30²)+(3/2)(3²+6²+9²+…+30²)
=(-1/2) ×30×31×61×(1/6)+(3/2) ×9×(1²+2²+3²+...+10²)
=（-5×31×61/2）+(27/2) ×5×7×11
=470
∴S30=470