f(x)=sin[π/3(x+1)]-根号3cos[π/3(x+1)],则f(1)+f(2)+f(3)+……+f(200

∵f(x)=sin[π(x+1)/3]-√3cos[π(x+1)/3]
=sin(πx/3+π/3)-√3cos(πx/3+π/3)
=2sin(πx/3+π/3-π/3)
=2sin(πx/3)
∴f(x)周期T=2π/(π/3)=6
又f(1)+f(2)+f(3)+f(4)+f(5)+f(6)=0
且2008=334×6+4
故f(1)+f(2)+f(3)+……+f(2008)=f(1)+f(2)+f(3)+f(4)=2√3