既然是二次函数,那就设f（x）=ax²+bx+cf（x+1）=a（x+1）²+b（x+1）+c,f（x-1）=a（x-1）²+b（x-1）+cf(x+1)+f(x-1)=2ax²+2bx+2a+2c于是2ax²+2bx+2a+2c=2x^2-4x2a=2,2b=-4,2a+2c=0得a=1.b=...