已知数列{an}的通项公式an=ncos(nπ/3),其前n项和为Sn,则S2014等于
问题描述：
已知数列{an}的通项公式an=ncos(nπ/3),其前n项和为Sn,则S2014等于
答
函数f(n) = cos(nπ/3)的最小正周期是2π/(π/3) = 6 ； 由已知a n = ncos(nπ/3) ,计算可得：a 1 = cos(π/3) =1*(1/2) = 1/2 ； a 2 = 2cos(2π/3) = 2*(-1/2) = -1 ； a 3 = 3cos(3π/3) =3*(-1) = -3 ； a 4 = 4...