用数学归纳法证明

数学归纳法证明x^(2n

(1) n=1时 原式=x+y 显然能被x+y整除 (2) 设n=k时结论成立 则x^(2k-1)+y^(2k-1)能被x=y整除 (3) 当n=k+1时 原式=x^[2(k+1)-1]+y^[2(k+1)-1] =x^(2k+1)+y(2k+1) =x^(2k-1)*x^2+y^(2k-1)*y^2 =x^(2k-1)*x^2-x^(2k-1)*y^2+x^(2k-1)*y^2+y^(2k-1)*y^2 =[x^(2k-1)*x^2-x^(2k-1)*y^2]+[x^(2k-1)*y^2+y^(2k-1)*y^2] =[x^(2k-1)*(x^2-y^2)]+{[x^(2k-1)+y^(...全部

  (1) n=1时 原式=x+y 显然能被x+y整除 (2) 设n=k时结论成立 则x^(2k-1)+y^(2k-1)能被x=y整除 (3) 当n=k+1时 原式=x^[2(k+1)-1]+y^[2(k+1)-1] =x^(2k+1)+y(2k+1) =x^(2k-1)*x^2+y^(2k-1)*y^2 =x^(2k-1)*x^2-x^(2k-1)*y^2+x^(2k-1)*y^2+y^(2k-1)*y^2 =[x^(2k-1)*x^2-x^(2k-1)*y^2]+[x^(2k-1)*y^2+y^(2k-1)*y^2] =[x^(2k-1)*(x^2-y^2)]+{[x^(2k-1)+y^(2k-1)]*y^2} =[x^(2k-1)*(x+y)*(x-y)]+{[x^(2k-1)+y^(2k-1)]*y^2} x^(2k-1)*(x+y)*(x-y)显然能被x+y整除 由(2)知[x^(2k-1)+y^(2k-1)]*y^2能被x+y整除 所以它们的和一定能被x+y整除 所以n=k+1时成立 证毕 。
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