取y=0得4f(x)f(0)=2f(x)得f(0)=1/2;
取y=1得 f(x)=f(x+1)+f(x-1)
进一步得 f(x+1)=f(x)+f(x+2)
两式相加得 f(x-1)+f(x+2)=0,
进一步得 f(x+2)+f(x+5)=0,
以上两式相减得f(x-1)=f(x+5),
进一步得 f(x)=f(x+6)
所以T=6;
F(2010)=F(0+335*6)=F(0)=1/2.
取y=0得4f(x)f(0)=2f(x)得f(0)=1/2;
取y=1得 f(x)=f(x+1)+f(x-1)
进一步得 f(x+1)=f(x)+f(x+2)
两式相加得 f(x-1)+f(x+2)=0,
进一步得 f(x+2)+f(x+5)=0,
以上两式相减得f(x-1)=f(x+5),
进一步得 f(x)=f(x+6)
所以T=6;
F(2010)=F(0+335*6)=F(0)=1/2.