用数学归纳法证明x^2n-1+y^2n-1能被x+y整除

用数学归纳法证明x^2n-1+y^2n-1能被x+y整除

明天考试，

问答/416℃/2024-04-12 20:18:32

优质解答：

当n=1时

x^(2n-1)+y^(2n-1)

=x+y

(x+y)/(x+y)=1

能被x+y整除.

假设当n=k(k为整数,且k>=2)时,x^(2k-1)+y^(2k-1)能被x+y整除,

则当n=k=1时

令x^(2k-1)+y^(2k-1)=A(x+y)

则x^(2k-1)=A(x+y)-y^(2k-1)

x^[2(k+1)-1]+y^[2(k+1)-1]

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

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

=x^2*[A(x+y)-y^(2k-1)]+y^2*y^(2k-1)

=x^2*A(x+y)-x^2y^(2k-1)+y^2*y^(2k-1)

=x^2*A(x+y)+(y^2-x^2)*y^(2k-1)

=x^2*A(x+y)+(x+y)(y-x)*y^(2k-1)

两项中均含x+y

[x^2*A(x+y)+(x+y)(y-x)*y^(2k-1)]/(x+y)

=Ax^2+(y-x)*y^(2k-1)为整数

能被x+y整除.

综上,x^(2n-1)+y^(2n-1)能被x+y整除

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