Then f (x+3) = f [(x+3/2)+3/2] =-f (x+3/2) = f (x), so the function period is 3.

Because the image of the function f(x) is symmetrical about the point (-3/4,0), then f(x)+f(-3/2-x)=0.

And f(x+3/2)=-f(x), so f(-3/2-x) =f(x+3/2), that is, f(t)=f(-t), and the function is even.

∴f(2)=f(2-3)=f(- 1)=f( 1)= 1,f(3)= f(3-3)= f(0)=-2

So f (1)+f (2)+f (3) =1+1-2 = 0.

f( 1)+f(2)+f(3)+…+f(20 1 1)

=[f( 1)+f(2)+f(3)]×670+f(20 1 1)

=[f( 1)+f(2)+f(3)]×670+f( 1)

=0+ 1= 1.