已知 f(n)=sin nπ/3，则f(1)+f(2)+......+f(2012)的值为

解答： 从周期去入手， f(n)=sin nπ/3，周期T=2π/(π/3)=6 ∵ f(1)=sin(π/3)=√3/2 f(2)=sin(2π/3)=√3/2 f(3)=sin(π)=0 f(4)=sin(4π/3)=-√3/2 f(5)=sin(5π/3)=-√3/2 f(6)=sin(2π)=0 ∴ 连续六项的和为0 又2012=6*335+2 ∴ f(