數列{a n }的通項a n =n²;(cos²;nπ/3-sin²;nπ/3),其前n項和為Sn,則S30為?

數列{a n }的通項a n =n²;(cos²;nπ/3-sin²;nπ/3),其前n項和為Sn,則S30為?

an =n²;(cos²;nπ/3-sin²;nπ/3)=n²;*cos(2nπ/3)對於函數f(n)=cos(2nπ/3),週期T=2π/(2π/3)=3則可知:當n=3k+1,k屬於N時,an=(3k+1)²;*cos[2(3k+1)π/3]=(3k+1)²;*cos(2kπ+2π/3)=(3k+1)…