数列{a n } 的通项a n =n²(cos²nπ/3-sin²nπ/3),其前n项和为Sn,则S30为?

题目

数列{a n } 的通项a n =n²(cos²nπ/3-sin²nπ/3),其前n项和为Sn,则S30为?

答案

an =n²(cos²nπ/3-sin²nπ/3)=n²*cos(2nπ/3)对于函数f(n)=cos(2nπ/3),周期T=2π/(2π/3)=3则可知：当n=3k+1,k属于N时,an=(3k+1)²*cos[2(3k+1)π/3]=(3k+1)²*cos(2kπ+2π/3)=(3k+1)...