Given { ((a_0 =1)),((a_(n+1) =(1/2)(3a_n +(√(5a_n ^2 −4)) ))) :} ∀n≥0 , n∈I find a


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 184280 by mr W last updated on 04/Jan/23
$Given { ((a_0 =1)),((a_(n+1) =(1/2)(3a_n +(√(5a_n ^2 −4)) ))) :} ∀n≥0 , n∈I find a_n .$
$G i v e n {\begin{cases} a_{0} = 1 \\ a_{n + 1} = \frac{1}{2} (3 a_{n} + \sqrt{5 a_{n}^{2} - 4}) \end{cases}$ $\forall n ⩾ 0, n \in I$ $f i n d a_{n} .$
Commented by mr W last updated on 04/Jan/23

$u n s o l v e d o l d q u e s t i o n$
Commented by mr W last updated on 04/Jan/23

$i f o u n d f i n a l l y a s o l u t i o n :$ $a_{n} = \frac{2}{\sqrt{5}} \cosh [(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}]$ $w h i c h d e l i v e r s :$ $a_{0} = 1$ $a_{1} = 2$ $a_{2} = 5$ $a_{3} = 13$ $a_{4} = 34$ $a_{5} = 89$ $. . .$
Commented by mr W last updated on 04/Jan/23

$m o r e o t h e r s o l u t i o n s ?$
Commented by SEKRET last updated on 04/Jan/23

$a_{n} = \frac{a_{n - 1} + a_{n + 1}}{3}$
Commented by SEKRET last updated on 04/Jan/23

$a_{0} = 1$ $a_{1} = 2 \cdot 1 + 0$ $a_{2} = 2 + (2 + 1)$ $a_{3} = 5 + (5 + 2 + 1)$ $a_{4} = 13 + (13 + 5 + 2 + 1)$ $a_{5} = 34 + (34 + 13 + 5 + 2 + 1)$ $. . . . . . . . . . .$ $a_{n} = a_{n - 1} + \underset{k = 0}{\sum^{n - 1}} a_{k}$
Commented by SEKRET last updated on 04/Jan/23

$a_{n} = \frac{2}{\sqrt{5}} \cosh ((2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2})$ $a_{0} = 1 a_{1} = 2 a_{3} = 13 a_{4} = 34 a_{5} = 89$ $a_{6} = 233 a_{7} = 610 a_{8} = 1597$
Commented by SEKRET last updated on 04/Jan/23

$how did you find it$
Commented by mr W last updated on 04/Jan/23

$t h a n k s!$ $b u t t h e q u e s t i o n a s k s f o r a_{n} i n$ $e x p l i c i t f o r m . r e c u r s i v e f o r m i s$ $a l r e a d y g i v e n i n t h e q u e s t i o n .$
Commented by SEKRET last updated on 04/Jan/23

$right$
Commented by Frix last updated on 04/Jan/23

${It}^{'} s each 2 nd Fibonacci - Number \Rightarrow$ $a_{n} = \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{2 n + 1} - {(\frac{1 - \sqrt{5}}{2})}^{2 n + 1})$
Commented by mr W last updated on 04/Jan/23

$h o w d i d y o u g e t t h i s ?$
Commented by manxsol last updated on 05/Jan/23

$t r u \bar{e}, g o o d o b s e r v a t i o n S r . F r i x$ $\begin{array}{ccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ f_{n} & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 \\ a_{n} & 2 & 5 & 13 & 34 & 89 \end{array}$ $a_{n} = f_{2 n - 2} a_{1} = 1$ $t h i s s o l u t i o n i s v a l i d ?$ $a_{n} = f_{2 n - 2}$
Commented by mr W last updated on 05/Jan/23

$s o f a r a s i t i s n o t d e v i a t e d f r o m t h e$ $g i v e n e q u a t i o n a_{n + 1} = \frac{1}{2} (3 a_{n} + \sqrt{5 a_{n}^{2} - 4}),$ $i {d o n}^{'} t t h i n k i t i s m a t h e m a t i c a l l y$ $s t r i c t, e v e n t h o u g h i t i s c o r r e c t .$ $t h e q u e s t i o n i s n o t :$ $g i v e n : 1, 2, 5, 13, 34, . . . f i n d a_{n} = ? .$ $a n d e v e n s o, t h e r e a r e i n f i n i t e$ $s o l u t i o n s f o r a_{n} a s w e k n o w .$
	Answered by mr W last updated on 05/Jan/23

	$a_{n + 1} = \frac{1}{2} (3 a_{n} + \sqrt{5 a_{n}^{2} - 4})$ $2 a_{n + 1} - 3 a_{n} = \sqrt{5 a_{n}^{2} - 4}$ $a_{n + 1}^{2} - 3 a_{n + 1} a_{n} + a_{n}^{2} + 1 = 0$ ${(a_{n + 1} - a_{n})}^{2} + 1 = a_{n + 1} a_{n}$ $l e t$ $a_{n + 1} + a_{n} = u$ $a_{n + 1} - a_{n} = v$ $\Rightarrow u^{2} - 5 v^{2} = 4$ ${(\frac{u}{2})}^{2} - {(\frac{\sqrt{5} v}{2})}^{2} = 1$ $\cosh^{2} x_{n} - \sinh^{2} x_{n} = 1$ $\frac{u}{2} = \cosh x_{n} \Rightarrow u = 2 \cosh x_{n}$ $\frac{\sqrt{5} v}{2} = \sinh x_{n} \Rightarrow v = \frac{2 \sinh x_{n}}{\sqrt{5}}$ $a_{n} = \frac{1}{2} (u - v) = \cosh x_{n} - \frac{\sinh x_{n}}{\sqrt{5}}$ $a_{n} = \frac{2}{\sqrt{5}} (\frac{\sqrt{5}}{2} \cosh x_{n} - \frac{1}{2} \sinh x_{n})$ $a_{n} = \frac{2}{\sqrt{5}} (\cosh x_{n} \cosh α - \sinh x_{n} \sinh α) = k \cosh (x_{n} - α)$ $a_{n} = \frac{2}{\sqrt{5}} \cosh (x_{n} - α) w i t h α = \cosh^{- 1} \frac{\sqrt{5}}{2}$ $a_{n + 1} = \frac{1}{2} (u + v) = \cosh x_{n} + \frac{\sinh x_{n}}{\sqrt{5}}$ $a_{n + 1} = \frac{2}{\sqrt{5}} (\frac{\sqrt{5}}{2} \cosh x_{n} + \frac{1}{2} \sinh x_{n})$ $a_{n + 1} = \frac{2}{\sqrt{5}} (\cosh x_{n} \cosh α + \sinh x_{n} \sinh α)$ $a_{n + 1} = \frac{2}{\sqrt{5}} \cosh (x_{n} + α)$ $o n t h e o t h e r s i d e :$ $a_{n + 1} = \frac{2}{\sqrt{5}} \cosh (x_{n + 1} - α)$ $\Rightarrow \cosh (x_{n + 1} - α) = \cosh (x_{n} + α)$ $\Rightarrow x_{n + 1} - α = x_{n} + α$ $\Rightarrow x_{n + 1} = x_{n} + 2 α \leftarrow A . P . w i t h c . d . = 2 α$ $\Rightarrow x_{n} = x_{0} + 2 n α$ $a_{n} = \frac{2}{\sqrt{5}} \cosh (x_{0} + 2 n α - α)$ $\Rightarrow a_{n} = \frac{2}{\sqrt{5}} \cosh [x_{0} + (2 n - 1) \cosh^{- 1} \frac{\sqrt{5}}{2}]$ $a_{0} = \frac{2}{\sqrt{5}} \cosh (x_{0} - \cosh^{- 1} \frac{\sqrt{5}}{2}) = 1$ $\Rightarrow x_{0} = 2 \cosh^{- 1} \frac{\sqrt{5}}{2}$ $\Rightarrow a_{n} = \frac{2}{\sqrt{5}} \cosh [(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}]$ $a d d i t i o n a l l y :$ $\cosh^{- 1} \frac{\sqrt{5}}{2} = \ln (\frac{\sqrt{5}}{2} + \sqrt{{(\frac{\sqrt{5}}{2})}^{2} - 1}) = \ln \frac{\sqrt{5} + 1}{2}$ $(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2} = \ln {(\frac{\sqrt{5} + 1}{2})}^{(2 n + 1)}$ $e^{- (2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}} = {(\frac{\sqrt{5} + 1}{2})}^{- (2 n + 1)} = {(\frac{\sqrt{5} - 1}{2})}^{(2 n + 1)}$ $a_{n} = \frac{2}{\sqrt{5}} \cosh [(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}]$ $a_{n} = \frac{1}{\sqrt{5}} [e^{(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}} + e^{- (2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}}]$ $\Rightarrow a_{n} = \frac{1}{\sqrt{5}} [{(\frac{\sqrt{5} + 1}{2})}^{(2 n + 1)} + {(\frac{\sqrt{5} - 1}{2})}^{(2 n + 1)}]$
	Commented by SEKRET last updated on 05/Jan/23

	$beatiful solution$
	Commented by Frix last updated on 05/Jan/23

	$Remember that \cosh α = \frac{e^{α} + e^{- α}}{2}$ $a_{n} = \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{2 n + 1} - {(\frac{1 - \sqrt{5}}{2})}^{2 n + 1}) =$ $= \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{2 b + 1} + {(\frac{2}{1 + \sqrt{5}})}^{2 n + 1}) =$ $= \frac{1}{\sqrt{5}} (t^{x} + t^{- x}) =$ $= \frac{1}{\sqrt{5}} (e^{x \ln t} + e^{- x \ln t}) =$ $= \frac{2}{\sqrt{5}} \cosh (x \ln t) =$ $= \frac{2}{\sqrt{5}} \cosh ((2 n + 1) \ln \frac{1 + \sqrt{5}}{2}) =$ $\ln (x + \sqrt{x^{2} - 1}) = \cosh^{- 1} x$ $\ln \frac{1 + \sqrt{5}}{2} = \ln (\frac{\sqrt{5}}{2} + \sqrt{\frac{5}{4} - 1}) = \cosh^{- 1} \frac{\sqrt{5}}{2}$ $= \frac{2}{\sqrt{5}} \cosh ((2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2})$ $. . . . which is your result$
	Commented by mr W last updated on 05/Jan/23

	$t h a n k s s i r!$ $i^{'} v e a l s o a d d e d t h e d e d u c t i o n t o$ $a_{n} = \frac{1}{\sqrt{5}} [{(\frac{\sqrt{5} + 1}{2})}^{(2 n + 1)} + {(\frac{\sqrt{5} - 1}{2})}^{(2 n + 1)}] .$ $s e e a b o v e .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum