已知f(n)=sin(nπ/6),则f（1）+f（2）+f（3）+…+f（2012）=．

f(1)=sin(π/6)=1/2 f(2)=sin(π/3)=√3/2 f(3)=sin(π/2)=1 f(4)=sin(2π/3)=√3/2 f(5)=sin(5π/6)=1/2 f(6)=sin(π)=0 f(7)=sin(7π/6)=-1/2 f(8)=sin(4π/3)=-√3/2 f(9)=sin(3π/2)=-1 f(10)=sin(5π/3)=-√3