Question

若n∈N,求证:xⁿ⁺¹+(x+1)^2n-1能被x²+x+1整除.

Answer 1

证明:(1)当n=1时,命题显然成立.

(2)设当n=k时,x^k+1+(x+1)^2k-1能被x²+x+1整除.

法1:(添项)当n=k+1时,

x^k+2+(x+1)^2k+1=(x+1)²(x+1)^2k-1+x^k+2+(x+1)²x^k+1-(x+1)²x^k+1

=(x+1)²［(x+1)^2k-1+x^k+1］-(x²+x+1)x^k+1,

而上面各项都能被x²+x+1整除,即n=k+1时成立.

法2:(拆项)当n=k+1时

x^k+2+(x+1)^2k+1=(x+1)²(x+1)^2k-1+x^k+2=(x²+x+1)(x+1)^2k-1+x［(x+1)^2k-1+x^k+1］,

以上各项都能被x²+x+1整除,即n=k+1时成立.

由(1)(2)命题得证.