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Question Number 143519 by tugu last updated on 15/Jun/21

T h e f i r s t t w o t e r m s o f t h e {a_{n}} s e r i e s a r e d e f i n d a s a_{n} = a_{n - 1} + a_{n - 2} f o r t h e g e n e r a l t e r m

a_{1} = 5, a_{2} = 8 a n d n ⩾ 3 .

s i n c e t h e L = l i \underset{n \to \infty}{m} \frac{a_{n + 1}}{a_{n}} w h a t i s t h e v a l u e o f L

Answered by mathmax by abdo last updated on 15/Jun/21

a_{n} = a_{n - 1} + a_{n - 2} \Rightarrow a_{n + 2} = a_{n + 1} + a_{n} \Rightarrow a_{n + 2} - a_{n + 1} - a_{n} = 0

(ec) \to x^{2} - x - 1 = 0 \to Δ = 1 + 4 = 5 \Rightarrow x_{1} = \frac{1 + \sqrt{5}}{2} and x_{2} = \frac{1 - \sqrt{5}}{2}

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

a_{1} = 5 \Rightarrow α (\frac{1 + \sqrt{5}}{2}) + β (\frac{1 - \sqrt{5}}{2}) = 5

a_{2} = 8 \Rightarrow α {(\frac{1 + \sqrt{5}}{2})}^{2} + β {(\frac{1 - \sqrt{5}}{2})}^{2} = 8 we get the system

{\begin{cases} \frac{1 + \sqrt{5}}{2} α + \frac{1 - \sqrt{5}}{2} β = 5 \\ {(\frac{1 + \sqrt{5}}{2})}^{2} α + {(\frac{1 - \sqrt{5}}{2})}^{2} β = 8 \end{cases}

Δ_{s} = \frac{(1 + \sqrt{5}) {(1 - \sqrt{5})}^{2}}{8} - \frac{(1 - \sqrt{5}) {(1 + \sqrt{5})}^{2}}{8}

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

Δ_{α} = | \begin{matrix} 5 \frac{1 - \sqrt{5}}{2} \\ 8 {(\frac{1 - \sqrt{5}}{2})}^{2} \end{matrix} | = \frac{5}{4} {(1 - \sqrt{5})}^{2} - 4 (1 - \sqrt{5})

= \frac{5}{4} (6 - 2 \sqrt{5}) - 4 + 4 \sqrt{5} = \frac{5}{2} (3 - \sqrt{5}) - 4 + 4 \sqrt{5} = \frac{15}{2} - 4 - \frac{5 \sqrt{5}}{2} + 4 \sqrt{5}

= \frac{7}{2} + \frac{3}{2} \sqrt{5} \Rightarrow α = \frac{7 + 3 \sqrt{5}}{2 \sqrt{5}}

Δ_{β} = | \begin{matrix} \frac{1 + \sqrt{5}}{2} 5 \\ {(\frac{1 + \sqrt{5}}{2})}^{2} 8 \end{matrix} | = 4 (1 + \sqrt{5}) - \frac{5}{4} {(1 + \sqrt{5})}^{2}

= 4 + 4 \sqrt{5} - \frac{5}{4} (6 + 2 \sqrt{5}) = 4 + 4 \sqrt{5} - \frac{5}{2} (3 + \sqrt{5}) = 4 + 4 \sqrt{5} - \frac{15}{2} - \frac{5 \sqrt{5}}{2}

= - \frac{7}{2} + \frac{3}{2} \sqrt{5} \Rightarrow β = \frac{- 7 + 3 \sqrt{5}}{2 \sqrt{5}} \Rightarrow

a_{n} = (\frac{7 + 3 \sqrt{5}}{2 \sqrt{5}}) {(\frac{1 + \sqrt{5}}{2})}^{n} + (\frac{- 7 + 3 \sqrt{5}}{2 \sqrt{5}}) {(\frac{1 - \sqrt{5}}{2})}^{n}

\Rightarrow a_{n} = \frac{(7 + 3 \sqrt{5}) {(1 + \sqrt{5})}^{n}}{2^{n + 1} (\sqrt{5})} + \frac{(- 7 + 3 \sqrt{5}) {(1 - \sqrt{5})}^{n}}{2^{n + 1} (\sqrt{5})}

= \frac{1}{(\sqrt{5}) 2^{n + 1}} {(7 + 3 \sqrt{5}) {(1 + \sqrt{5})}^{n} - (7 - 3 \sqrt{5}) {(1 - \sqrt{5})}^{n}}

rest to calculate \frac{a_{n + 1}}{a_{n}} \dots . be continued \dots

Answered by mr W last updated on 15/Jun/21

x^{2} - x - 1 = 0

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

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

a_{0} = a_{2} - a_{1} = 8 - 5 = 3 = A + B \dots (i)

a_{1} = 5 = A (\frac{1 + \sqrt{5}}{2}) + B (\frac{1 - \sqrt{5}}{2}) \dots (i i)

\Rightarrow A = \frac{15 + 7 \sqrt{5}}{10}

\Rightarrow B = \frac{15 - 7 \sqrt{5}}{10}

a_{n} = (\frac{15 + 7 \sqrt{5}}{10}) {(\frac{1 + \sqrt{5}}{2})}^{n} + (\frac{15 - 7 \sqrt{5}}{10}) {(\frac{1 - \sqrt{5}}{2})}^{n}

a_{n} = {(\frac{1 + \sqrt{5}}{2})}^{n} [(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n}]

\frac{a_{n + 1}}{a_{n}} = (\frac{1 + \sqrt{5}}{2}) \times \frac{(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n + 1}}{(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) {(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})}^{n}}

\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = (\frac{1 + \sqrt{5}}{2}) \times \frac{(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) \times 0}{(\frac{15 + 7 \sqrt{5}}{10}) + (\frac{15 - 7 \sqrt{5}}{10}) \times 0} = \frac{1 + \sqrt{5}}{2} = φ

Answered by mr W last updated on 15/Jun/21

a_{n + 1} = a_{n} + a_{n - 1}

\frac{a_{n + 1}}{a_{n}} = 1 + \frac{1}{\frac{a_{n}}{a_{n - 1}}}

\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 1 + \frac{1}{\lim_{n \to \infty} \frac{a_{n}}{a_{n - 1}}}

L = 1 + \frac{1}{L} > 1

L^{2} - L - 1 = 0

\Rightarrow L = \frac{1 + \sqrt{5}}{2}

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