If-the-roots-of-the-equation-x-2-x-1-0-are-and-provided-that-x-n-n-n-Find-x-16-

Question Number 107812 by I want to learn more last updated on 12/Aug/20

If the roots of the equation x^{2} - x - 1 = 0 are α and β,

provided that x_{n} = α^{n} + β^{n} . Find x_{16} .

Answered by mr W last updated on 12/Aug/20

α + β = 1

α β = - 1

α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = 1 - 2 (- 1) = 3

α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 {(α β)}^{2} = 3^{2} - 2 = 7

α^{8} + β^{8} = {(α^{4} + β^{4})}^{2} - 2 {(α β)}^{4} = 7^{2} - 2 = 47

α^{16} + β^{16} = {(α^{8} + β^{8})}^{2} - 2 {(α β)}^{8} = 47^{2} - 2 = 2207

Commented by I want to learn more last updated on 12/Aug/20

I really appreciate sir .

Sir is there a way to solve higher power like x_{91} ? ?

Commented by mr W last updated on 12/Aug/20

α, β = \frac{1 \pm \sqrt{5}}{2}

α^{n} + β^{n} = {(\frac{1 + \sqrt{5}}{2})}^{n} + {(\frac{1 - \sqrt{5}}{2})}^{n}

Commented by I want to learn more last updated on 13/Aug/20

Thanks sir

Answered by hgrocks last updated on 12/Aug/20

x^{2} = x + 1

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

α^{n + 1} = α^{n} + α^{n - 1} (1)

β^{n + 1} = β^{n} + β^{n - 1} (2)

(1) + (2) \to

x_{n + 1} = x_{n} + x_{n - 1}

x_{0} = 2

x_{1} = α + β = 1

x_{2} = 2 + 1 = 3

x_{3} = 1 + 3 = 4

Recursive Sequence :

2, 1, 3, 4, 7, 11 \dots \dots \dots \dots

Commented by I want to learn more last updated on 13/Aug/20

Thanks sir .