题目内容
求证:二项式x2n-y2n (n∈N*)能被x+y整除.
证明略
(1)当n=1时,x2-y2=(x+y)(x-y),
能被x+y整除,命题成立.
(2)假设当n=k(k≥1,k∈N*)时,x2k-y2k能被x+y整除,
那么当n=k+1时,
x2k+2-y2k+2=x2·x2k-y2·y2k
=x2x2k-x2y2k+x2y2k-y2y2k
=x2(x2k-y2k)+y2k(x2-y2),
显然x2k+2-y2k+2能被x+y整除,
即当n=k+1时命题成立.
由(1)(2)知,对任意的正整数n命题均成立.
能被x+y整除,命题成立.
(2)假设当n=k(k≥1,k∈N*)时,x2k-y2k能被x+y整除,
那么当n=k+1时,
x2k+2-y2k+2=x2·x2k-y2·y2k
=x2x2k-x2y2k+x2y2k-y2y2k
=x2(x2k-y2k)+y2k(x2-y2),
显然x2k+2-y2k+2能被x+y整除,
即当n=k+1时命题成立.
由(1)(2)知,对任意的正整数n命题均成立.
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