Question

用数学归纳法证明当n是正奇数时,xⁿ+yⁿ能被x+y整除.

      

Answer 1

证明:(1)当n=1时,xⁿ+yⁿ=x+y能被x+y整除.?

       (2)假设n=2k-1时命题成立,即(x^2k-1+y^2k-1)能被x+y整除,那么n=2k+1时有x^2k+1+y^2k+1=x^2k-1·x²+y^2k-1·y²?

       =x^2k-1·x²+x^2k-1·y²-x^2k-1·y²+y^2k-1·y²

       =x^2k-1(x²-y²)+(x^2k-1+y^2k-1)·y²?

       =x^2k-1(x+y)(x-y)+(x^2k-1+y^2k-1)·y².?

       ∵x^2k-1(x+y)(x-y)与(x^2k-1+y^2k-1)y²都能被x+y整除,∴n=2k+1时命题成立.?

       综合(1)(2)可知对一切正奇数n,xⁿ+yⁿ都能被x+y整除?.