递推式说明相邻差相等：fn+2(x)−fn+1(x)=fn+1(x)−fn(x)f_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_n(x)fn+2(x)−fn+1(x)=fn+1(x)−fn(x)。从 f1(x)−f0(x)=f(x)−xf_1(x)-f_0(x)=f(x)-xf1(x)−f0(x)=f(x)−x 开始，求和 nnn 个相同差，得 fn(x)−x=n(f(x)−x)f_n(x)-x=n(f(x)-x)fn(x)−x=n(f(x)−x)。