已知f（x)=sin[π/3（x+1）]-abs（3）cos[π/3（x+1）] 则f(1)+f(2)+.+f(2006)=

f(x)=2(1/2sin[π/3（x+1)])]-√3/2cos[π/3（x+1)])
=2sin(πx/3)
则f(6k) = 0,f(6k+1) = f(6k+2) = √3
f(6k+3) = 0,f(6k+4) = f(6k+5) = -√3
连续6个相加为0,由 2006 = 334*6 + 2
∴原式 = f(2005)+f(2006) = f(1)+f(2) = 2√3