Question

已知定义在R上的函数f(x)满足f(x)=-f(x+),且f(-2)=f(-1)=-1,f(0)=2,则f(1)+f(2)+…+f(2 005)+f(2 006)等于(    )

A.-2             B.-1             C.0            D.1

Answer 1

解析:∵f(x)=-f(x+)=-［-f(x++)］=f(x+3),

    ∴f(x)的周期为3.

    又f(1)=f(-2+3)=f(-2)=-1,f(2)=f(-1+3)=f(-1)=-1,f(3)=f(0+3)=f(0)=2,

    从而f(1)+f(2)+f(3)=0.故f(1)+f(2)+…+f(2 005)+f(2 006)=f(2 005)+f(2 006)=f(3×668+1)+f(3×668+2)=f(1)+f(2)=-2.选A.

答案:A