It is known that f (x) is an odd function defined on R, and f (1) = 1. If the image of F (x) is shifted one unit to the right, the image of an even function is obtained, Then f (1) + F (2) + F (3) f (4) + F (2011) is equal to

It is known that f (x) is an odd function defined on R, and f (1) = 1. If the image of F (x) is shifted one unit to the right, the image of an even function is obtained, Then f (1) + F (2) + F (3) f (4) + F (2011) is equal to

∵ f (x) is an even function on R,
The image is symmetric about the y-axis, that is, the function has the axis of symmetry x = 0, f (x) = f (- x) 1
If the image of F (x) is shifted one unit to the right, the image of an odd function is obtained,
The function f (x) has the center of symmetry (- 1,0) before moving to the right, that is, f (- 1) = 0, and f (- 1-x) = - f (- 1 + x) 2
The existence period of function f (x) is t = 4, f (2) = - 1, f (- 1) = 0,
We can deduce that f (- 1) = f (1) = 0, f (2) = - 1 = - f (0), f (3) = f (4-1) = 0,
f（-3）=f（3）=0,f（4）=f（0）=1,
So in a period, f (1) + F (2) + F (3) + F (4) = 0,
So f (1) + F (2) + F (3) + +f（2011）=f（1）+f（2）+f（3）=-1．
So choose a