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Question Number 121014 by mathocean1 last updated on 04/Nov/20

show by recurrence that

\forall n ⩾ 1,

a^{n} - b^{n} = (a - b) (a^{n - 1} + a^{n - 2} * b + \dots + {ab}^{n - 2} + b^{n - 1})

Answered by mathmax by abdo last updated on 04/Nov/20

let prove by recurence that x^{n} - 1 = (x - 1) (x^{n - 1} + x^{n - 2} + \dots + x + 1)

for x real n natural > 0

n = 1 we get x^{1} - 1 = (x - 1) (1) (relation true)

let suppose P_{n} true

x^{n + 1} - 1 = x x^{n} - 1 = x (x^{n} - 1 + 1) - 1

= x (x^{n} - 1) + x - 1 = x (x - 1) (x^{n - 1} + x^{n - 2} + \dots + x + 1) + x - 1

= (x - 1) (x^{n} + x^{n - 1} + \dots + x^{2} + x + 1) so P_{n + 1} is true

for x = \frac{a}{b} and b \neq 0 we get

{(\frac{a}{b})}^{n} - 1 = (\frac{a}{b} - 1) (\frac{a^{n - 1}}{b^{n - 1}} + \frac{a^{n - 2}}{b^{n - 2}} + \dots . . + \frac{a}{b} + 1) \Rightarrow

\frac{a^{n} - b^{n}}{b^{n}} = (\frac{a - b}{b}) (\frac{a^{n - 1}}{b^{n - 1}} + \frac{a^{n - 2}}{b^{n - 2}} + \dots . + \frac{a}{b} + 1) \Rightarrow

a^{n} - b^{n} = (a - b) b^{n - 1} (\frac{a^{n - 1}}{b^{n - 1}} + \frac{a^{n - 2}}{b^{n - 2}} + \dots . + \frac{a}{b} + 1) \Rightarrow

a^{n} - b^{n} = (a - b) (a^{n - 1} + a^{n - 2} b + \dots . . {ab}^{n - 2} + b^{n - 1})