n=2k+1时 y1^2-y2^2+y3^2-y4^2+...+y(2k-1)^2-y2k^ 2n=2k时 表达式与上式相同
解: f(x1,x2,x3) = x1^2+x2^2+2x3^2+2x1x2-2x1x3-2x2x3 = (x1+x2-x3)^2 + x3^2 = y1^2+y2^2. C = 1 1 -1 0 0 1 0 1 0 Y=CX
