Let-u-1-1-u-2-1-and-u-n-2-u-n-1-u-n-find-u-n-

Question Number 121832 by bemath last updated on 12/Nov/20

L e t {\begin{cases} u_{1} = 1 \\ u_{2} = 1 \end{cases} a n d u_{n + 2} = u_{n + 1} + u_{n}

f i n d u_{n} .

Commented by bemath last updated on 12/Nov/20

t h a n k y o u b o t h

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

q^{2} - q - 1 = 0

q = \frac{1 \pm \sqrt{5}}{2}

u_{n} = A {(\frac{1 + \sqrt{5}}{2})}^{n} + B {(\frac{1 - \sqrt{5}}{2})}^{n}

u_{0} = A + B = u_{2} - u_{1} = 1 - 1 = 0

\Rightarrow A = - B

u_{1} = A (\frac{1 + \sqrt{5}}{2}) + B (\frac{1 - \sqrt{5}}{2}) = 1

A [\frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2}] = 1

\Rightarrow A = \frac{1}{\sqrt{5}}

\Rightarrow B = - \frac{1}{\sqrt{5}}

\Rightarrow u_{n} = \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}]

Answered by Dwaipayan Shikari last updated on 12/Nov/20

u_{n + 2} = u_{n + 1} + u_{n}

r^{n + 2} = r^{n + 1} + r^{n}

r^{2} - r - 1 = 0 \Rightarrow r = \frac{1 \pm \sqrt{5}}{2}

u_{n} = Λ {(\frac{1 + \sqrt{5}}{2})}^{n} + Γ {(\frac{1 - \sqrt{5}}{2})}^{n}

u_{1} = \frac{Λ + Γ}{2} + \sqrt{5} \frac{Λ - Γ}{2} = 1

u_{2} = 3 \frac{Λ + Γ}{2} + \sqrt{5} \frac{Λ - Γ}{2} = 1 \Rightarrow Λ + Γ = 0 \Rightarrow Λ = - Γ

1 = \frac{\sqrt{5} (- 2 Γ)}{2} \Rightarrow Γ = - \frac{1}{\sqrt{5}} a n d Λ = \frac{1}{\sqrt{5}}

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

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