f(x+3)
=f[(x+3/2)+3/2]
=-f(x+3/2)
=f(x)
所以周期T=3
奇函數(shù)
f(3)=f(0)=0
f(2)
=f(1/2+3/2)
=-f(1/2)
f(1)=f(-1/2+3/2)
=-f(-1/2)
=f(1/2)
所以f(1)+f(2)+f(3)=0
T=3
所以f(4)+f(5)+f(6)=0
……
所以原式=0+0+……+f(2011)+f(2012)
=f(1)+f(2)
=0