Question

用数学归纳法证明n为正偶数时xⁿ-yⁿ能被x+y整除.

Answer 1

证明：①当n=2时，x²-y²=(x+y)(x-y),

即x²-y²能被x+y整除，显然命题成立.

②假设n=2k(k∈N^*)时，命题成立，即x^2k-y^2k能被x+y整除.

则当n=2k+2时，x^2k+2-y^2k+2=x²·x^2k-y²·y^2k=x²(x^2k-y^2k)+y^2k(x²-y²)

=x²(x^2k-y^2k)+y^2k(x+y)(x-y).

∵x²(x^2k-y^2k)、y^2k(x+y)(x-y)都能被x+y整除,

∴x^2k+2-y^2k+2能被x+y整除，即n=2k+2时命题成立.