✔ 最佳答案
When n = 3 , x + y = (x + y) (x³⁻¹ - x³⁻² y + y³⁻¹) is true.Assume xᵏ + yᵏ = (x + y) (xᵏ⁻¹ - xᵏ⁻² y + ... + yᵏ⁻¹) is true for some positive odd integer k > 1 , thenxᵏ⁺² + yᵏ⁺²
= x(xᵏ⁺¹) + yxᵏ⁺¹ + y(yᵏ⁺¹) + xyᵏ⁺¹ - (yxᵏ⁺¹ + xyᵏ⁺¹)
= (x + y) xᵏ⁺¹ + (x + y) yᵏ⁺¹ - xy (xᵏ + yᵏ)
= (x + y) (xᵏ⁺¹ + yᵏ⁺¹) - xy (xᵏ + yᵏ)
= (x + y) (xᵏ⁺¹ + yᵏ⁺¹) - xy (x + y) (xᵏ⁻¹ - xᵏ⁻² y + ... + yᵏ⁻¹)
= (x + y) (xᵏ⁺¹ + yᵏ⁺¹ - xy (xᵏ⁻¹ - xᵏ⁻² y + ... + yᵏ⁻¹))
= (x + y) (xᵏ⁺¹ - (xᵏy - xᵏ⁻¹ y² + ... + xyᵏ) + yᵏ⁺¹)
= (x + y) (xᵏ⁺¹ - xᵏy + xᵏ⁻¹ y² - ... - xyᵏ + yᵏ⁺¹)It is true for n = k+2 when it is true for n = k.
By the principle of Mathematical Induction, it is true for all positive odd integer n > 1.
2013-10-09 22:25:37 補充：
Correction:
The 1st line is 'When n = 3 , x³ + y³ = (x + y) (x³⁻¹ - x³⁻² y + y³⁻¹) is true.'
