Mathematical induction

2013-10-10 4:08 am
Using the principle of Mathematical Induction, prove that
x^n + y^n = (x + y)[x^n-1 - (x^n-2)y +....+ y^n-1]
更新1:

for all positive odd integer.

更新2:

for all positive odd integer n > 1.

回答 (3)

2013-10-10 6:20 am
✔ 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.'
2013-10-10 7:22 am
☂雨後晴空☀ == 彩虹部隊 ???

2013-10-10 00:08:18 補充:
OK。收到。

^___^
2013-10-10 6:28 am
好野呀!
彩虹部隊出現啦~
我今次只夠快排第四~

^__^

2013-10-09 23:25:09 補充:
對唔住呀~
呢度有少少秘密係暫時未方便講的~

^___^


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