f(x)=sin[π/3(x+1)]-根号3cos[π/3(x+1)],则f(1)+f(2)+f(3)+……+f(2008)等于?

∵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)+…...