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Question Number 198152 by universe last updated on 11/Oct/23

a_{n + 2} = \sqrt{a_{n} \times a_{n + 1}} \forall n ⩾ 1, n \in N

and here a_{1} = α a n d a_{2} = β then

prove that \lim_{n \to \infty} a_{n + 2} = {(α \times β^{2})}^{1 / 3}

Answered by mr W last updated on 12/Oct/23

l e t b_{n} = \ln a_{n}

a_{n + 2} = \sqrt{a_{n + 1} \times a_{n}}

\ln a_{n + 2} = \frac{\ln a_{n + 1} + \ln a_{n}}{2}

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

\Rightarrow 2 b_{n + 2} - b_{n + 1} - b_{n} = 0 (r e c u r r e n c e r e l a t i o n)

2 p^{2} - p - 1 = 0 (c h a r a c t e r i s t i c e q u a t i o n)

(2 p + 1) (p - 1) = 0

\Rightarrow p = 1, - \frac{1}{2}

\Rightarrow b_{n} = A + B {(- \frac{1}{2})}^{n}

b_{1} = A + B (- \frac{1}{2}) = \ln a_{1} = \ln α \dots (i)

b_{2} = A + B {(- \frac{1}{2})}^{2} = \ln a_{2} = \ln β \dots (i i)

(i i) - (i) :

\frac{3 B}{4} = \ln β - \ln α

\Rightarrow B = \frac{4}{3} (\ln \frac{β}{α}) = \ln {(\frac{β}{α})}^{\frac{4}{3}}

\Rightarrow A = \frac{B}{2} + \ln α = \frac{2}{3} (\ln \frac{β}{α}) + \ln α = \ln {(α β^{2})}^{\frac{1}{3}}

\Rightarrow b_{n} = \ln {(α β^{2})}^{\frac{1}{3}} + {(- \frac{1}{2})}^{n} \ln {(\frac{β}{α})}^{\frac{4}{3}}

\Rightarrow b_{n} = \ln [{(α β^{2})}^{\frac{1}{3}} {(\frac{β}{α})}^{\frac{4}{3} {(- \frac{1}{2})}^{n}}] = \ln a_{n}

\Rightarrow a_{n} = {(α β^{2})}^{\frac{1}{3}} {(\frac{β}{α})}^{\frac{4}{3} {(- \frac{1}{2})}^{n}}

\Rightarrow \lim_{n \to \infty} a_{n} = {(α β^{2})}^{\frac{1}{3}} ✓

Commented by universe last updated on 12/Oct/23

t h a n k s s i r

Answered by Frix last updated on 12/Oct/23

a_{n + 2} = \sqrt{a_{n + 1} a_{n}}

a_{n} = e^{C_{1} {(- \frac{1}{2})}^{n} + C_{2}}

a_{1} = α \Leftrightarrow e^{- \frac{C_{1}}{2} + C_{2}} = α \Leftrightarrow - \frac{C_{1}}{2} + C_{2} = \ln α

a_{2} = β \Leftrightarrow e^{\frac{C_{1}}{4} + C_{2}} = β \Leftrightarrow \frac{C_{1}}{4} + C_{2} = \ln β

\Rightarrow

C_{1} = \frac{4}{3} \ln \frac{β}{α} \land C_{2} = \frac{1}{3} \ln α β^{2}

\Rightarrow

a_{n} = α^{\frac{1}{3} (1 - {(- 2)}^{2 - n})} β^{\frac{2}{3} (1 + {(- 2)}^{1 - n})}

\lim_{n \to \infty} a_{n} = α^{\frac{1}{3} (1 \pm 0)} β^{\frac{2}{3} (1 \mp 0)} = {(α β^{2})}^{\frac{1}{3}}

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