{ ((U_0 =1 et U_1 =2)),((U_(n+2) =(√(U_n U_(n+1) )))) :} determiner le terme generale et sa nature besoin d′aide avp


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Question Number 181323 by KONE last updated on 24/Nov/22
${ ((U_0 =1 et U_1 =2)),((U_(n+2) =(√(U_n U_(n+1) )))) :} determiner le terme generale et sa nature besoin d′aide avp$
${\begin{cases} U_{0} = 1 e t U_{1} = 2 \\ U_{n + 2} = \sqrt{U_{n} U_{n + 1}} \end{cases}$ $d e t e r m i n e r l e t e r m e g e n e r a l e e t s a n a t u r e$ $b e s o i n d^{'} a i d e a v p$
Commented by KONE last updated on 24/Nov/22

$∙ s v p$
	Answered by Frix last updated on 24/Nov/22

	$u_{0} = 2^{0}$ $u_{1} = 2^{1}$ $u_{2} = 2^{\frac{1}{2}}$ $u_{3} = 2^{\frac{3}{4}}$ $u_{4} = 2^{\frac{5}{8}}$ $. . .$ $let u_{n} = 2^{\frac{v_{n}}{2^{n - 1}}}$ $v_{0} = 0$ $v_{1} = 1$ $v_{2} = 1$ $v_{3} = 3$ $v_{4} = 5$ $v_{n} = 2 v_{n - 2} + v_{n - 1}$ $v_{n} = \frac{2^{n} - {(- 1)}^{n}}{3}$ $\Rightarrow$ $u_{n} = 2^{\frac{2^{n} - {(- 1)}^{n}}{3 \times 2^{n - 1}}} = 2^{\frac{2}{3} (1 - \frac{1}{{(- 2)}^{n}})}$ $\lim_{n \to \infty} u_{n} = 2^{\frac{2}{3}}$
	Commented by KONE last updated on 25/Nov/22

	$m e r c i$
	Answered by mr W last updated on 24/Nov/22

	$U_{n + 2} = \sqrt{U_{n + 1} U_{n}}$ $\ln U_{n + 2} = \frac{1}{2} (\ln U_{n + 1} + \ln U_{n})$ $l e t a_{n} = \ln U_{n}$ $2 a_{n + 2} - a_{n + 1} - a_{n} = 0$ $l e t a_{n} = {C p}^{n}$ $2 {C p}^{n + 2} - {C p}^{n + 1} - {C p}^{n} = 0$ ${C p}^{n} (2 p^{2} - p - 1) = 0$ $2 p^{2} - p - 1 = 0$ $(2 p + 1) (p - 1) = 0$ $\Rightarrow p = - \frac{1}{2} o r 1$ $\Rightarrow a_{n} = C \times {(- \frac{1}{2})}^{n} + D \times {(1)}^{n} = \frac{C}{{(- 2)}^{n}} + D$ $a_{0} = \ln U_{0} = \ln 1 = 0 = \frac{C}{{(- 2)}^{0}} + D$ $\Rightarrow C + D = 0 . . . (i)$ $a_{1} = \ln U_{1} = \ln 2 = \frac{C}{{(- 2)}^{1}} + D = - \frac{C}{2} + D$ $\Rightarrow - C + 2 D = 2 \ln 2 . . . (i i)$ $\Rightarrow D = \frac{2 \ln 2}{3} \Rightarrow C = - \frac{2 \ln 2}{3}$ $\Rightarrow a_{n} = \frac{2 \ln 2}{3} (1 - \frac{{(- 1)}^{n}}{2^{n}}) = \ln 2^{\frac{2}{3} (1 - \frac{{(- 1)}^{n}}{2^{n}})}$ $\Rightarrow U_{n} = e^{a_{n}} = 2^{\frac{2}{3} [1 - \frac{{(- 1)}^{n}}{2^{n}}]} ✓$ $\lim_{n \to \infty} U_{n} = 2^{\frac{2}{3}} = \sqrt[3]{4}$
	Commented by mnjuly1970 last updated on 24/Nov/22

	$s i r W (e x c u s e m e)$ $e^{a_{n}} = \sqrt[3]{4} . e^{1 - \frac{{(- 1)}^{n}}{2^{n}}} ? ? ?$
	Commented by mnjuly1970 last updated on 24/Nov/22

	$t h a n k s a l o t s i r . .$ $m y m i s t a k e$ $i n f a c t : a_{n} = \ln {(2)}^{\frac{2}{3} (1 - \frac{{(- 1)}^{n}}{2^{n}})}$
	Commented by mr W last updated on 24/Nov/22

	$n o!$ $e^{a_{n}} = {(\sqrt[3]{4})}^{1 - \frac{{(- 1)}^{n}}{2^{n}}}$
	Commented by mnjuly1970 last updated on 24/Nov/22

	Commented by KONE last updated on 25/Nov/22

	$m e r c i$
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