若函数fx=cosπx/3,（x属于整数）则f(0)+f(1)+f(2)+.f（2009）=

∵f(x)=cos(xπ/3)
∴f(0)=1
f(1)=cosπ/3=1/2
f(2)=cos2π/3=cos(π/2-π/6)
=-sin(π/6)
=-1/2
f(3)=cosπ=-1
f(4)=cos(4π/3)
=cos(π+π/3)
=-cosπ/3
=-1/2
f(5)=cos5π/3
=cos(2π-π/3)
=cosπ/3
=1/2
……
f(2009)=cos2009π/3
=cos(335*2π-π/3)
=cosπ/3π
=1/2
又2009+1=335*6
∴f(0)+f(1)+f(2)+f(3)+……+f(2009)=1+(1/2)+(-1/2)+(-1)+……+（1/2)
=0