若f(x)=sin(∏x/3),则f(1)+f(2)+f(3)+...+f(2003)=
问题描述：
若f(x)=sin(∏x/3),则f(1)+f(2)+f(3)+...+f(2003)=
答
当x=6k-5,(k∈N※,下同).则f(6k-5)=sin(2kπ-5π/3)=-sin(5π/3)=√3/2.f(6k-4)=sin(2kπ-4π/3)=√3/2f(6k-3)=sin(2kπ-3π/3)=0f(6k-2)=sin(2kπ-2π/3)=-√3/2f(6k-1)=sin(2kπ-π/3)=-√3/2f(6k)=sin2kπ=0.∴f(...