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

Question

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

Answer 1

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(1)+f(2)+.+f(2012)
=f(1)+f(2)+[f(3)+f(4)+.+f(2012)]
后面的6*355项可以6个6个组合,共有355组,每组和为0
=f(1)+f(2)+0
=√3/2+√3/2
=√3
即f(1)+f(2)+.+f(2012)的值为√3