記f(n)=x^n -nx + n-1,
n=1時(shí),f(1)=x-x=0顯然能被(x-1)^2整除.
設(shè)n=k時(shí),f(k)能被(x-1)^2整除,則當(dāng)n=k+1時(shí)
f(k+1)-f(k)=x^(k+1) - x^k -x +1 = x^k(x-1) -(x-1) = (x-1)(x^k-1)=(x-1)^2 * [1+x+...+x^(k-1)]
所以f(k+1)-f(k)能被(x-1)^2整除,再由歸納假設(shè)有f(k)能被(x-1)^2整除,所以f(k+1)能被(x-1)^2整除.
所以對(duì)任意自然數(shù)n,f(n)=x^n -nx + n-1能被(x-1)^2整除.