定义在R上的函数f(x)=log2(8-x),x0则f(2013)=( )

f(x+1)=f(x)-f(x-1)
=f(x-1)-f(x-2)-f(x-1)
=-f(x-2)
=-f(x-3)+f(x-4)
=-f(x-4)+f(x-5)+f(x-4)
=f(x-5)
=f(x-6)-f(x-7)
=f(x-7)-f(x-8)-f(x-7)
=-f(x-8)
=-f(x-9)+f(x-10)
=-f(x-10)+f(x-11)+f(x-10)
=f(x-11)
由上面规律得x+1-x+5=6 x-5-x+11=6可以得到
f(x)是以6为周期的函数
所以f(x)=f(x-6)=f(x-12)=.=f(x-6k)
所以f(2013)=f(2013-6k) 令k=385
f(2013)=f(2013-6*385)=f(3) =f(2)-f(1)=f(1)-f(0)-f(1)=-f(0)
=-log2 8=-3 选D