Question

已知定义在R上的函数f(x)满足f(x)=-f(x+3/2),且f(-2)=f(-1)=-1,f(0)=2,f(1)+f(2)+...+f(2009)=

Answer 1

因为f(x)=-f(x+3/2)=f(x+3),所以f(-2)=f(-2+3)=-1,f(-1）=f(-1+3）=-1，f(o)=f(0+3)=2,f(I)=f(1+3)=-1..........f(2006)=f(2006+3)=-2,（2009+1）/3=670没余数，f(0)+f(1)+f(3)=0,那可计算出f(1)+f(2)+f(3)+........f(2008)+f(2009)=-2

Answer 2

f(x)=-f(x+3/2)=-[-f(x+3)]=f(x+3)
f(x)为周期函数f（1)+f(2)+f(3)++f(1+n*3)+f(2+n*3)+f(3+n*3)

Answer 3

f(x)=-f(x+3/2)=-[-f(x+3)]=f(x+3)
f(1)+f(2)+...+f(2009)=669[f(-2)+f(-1)+f(0)]+f(-2)+f(-1)=-2