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

Question

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

Answer 1

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)=-√3/2
f(11)=sin(11π/6)=-1/2
f(12)=sin(2π)=0
所以f(n)是关于1/2,√3/2,1,√3/2,1/2,0,-1/2,-√3/2,-1,-√3/2,-1/2,0循环
f(1)+f(2)+.+f(12)=0
2012/12=167.8
f（1）+f（2）+f（3）+…+f（2012）
=0*167+1/2+√3/2+1+√3/2+1/2+0-1/2-√3/2
=1+√3/2+1/2
=√3/2+3/2