somebody has posted following question and then deleted it again. { ((u_(n+1) =((4u_n −9)/(u_n −2)))),((u_0 =5)) :} find u


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Question Number 214566 by mr W last updated on 12/Dec/24
$somebody has posted following question and then deleted it again. { ((u_(n+1) =((4u_n −9)/(u_n −2)))),((u_0 =5)) :} find u_n =? (or something like this)$
$s o m e b o d y h a s p o s t e d f o l l o w i n g$ $q u e s t i o n a n d t h e n d e l e t e d i t a g a i n .$ ${\begin{cases} u_{n + 1} = \frac{4 u_{n} - 9}{u_{n} - 2} \\ u_{0} = 5 \end{cases}$ $f i n d u_{n} = ? (o r s o m e t h i n g l i k e t h i s)$
	Answered by Hanuda354 last updated on 12/Dec/24
	${u_n } = { 5, ((11)/3) , ((17)/5) , ((23)/7) , …, 3 + (2/(2n+1)) } for n≥0$
	${u_{n}} = {5, \frac{11}{3}, \frac{17}{5}, \frac{23}{7}, \dots, 3 + \frac{2}{2 n + 1}}$ $for n ⩾ 0$
	Commented by mr W last updated on 12/Dec/24

	$p l e a s e s h a r e h o w y o u g e t t h i s .$
	Commented by Hanuda354 last updated on 13/Dec/24

	$Using iteration, we get :$ $u_{1} = \frac{4 (5) - 9}{5 - 2} = \frac{11}{3}$ $u_{2} = \frac{4 (\frac{11}{3}) - 9}{\frac{11}{3} - 2} = \frac{17}{5}$ $u_{3} = \frac{4 (\frac{17}{5}) - 9}{\frac{17}{5} - 2} = \frac{23}{7}$ $u_{4} = \frac{4 (\frac{23}{7}) - 9}{\frac{23}{7} - 2} = \frac{29}{9}$ $⋮$ $u_{n} = \frac{6 n + 5}{2 n + 1} = 3 + \frac{2}{2 n + 1}$ $By observing the pattern, we can conjecture a general$ $formula u_{n} :$ $u_{n} = 3 + \frac{2}{2 n + 1}$ $Proof by Induction :$ $Base case : u_{0} = 3 + \frac{2}{2 (0) + 1} = 5$ $Assume the formula holds for n = k, i . e, u_{k} = 3 + \frac{2}{2 k + 1} .$ $We need to prove it holds for n = k + 1 .$ $u_{k + 1} = \frac{4 u_{k} - 9}{u_{k} - 2} = \frac{4 (3 + \frac{2}{2 k + 1}) - 9}{3 + \frac{2}{2 k + 1} - 2}$ $Simplifying the expression, we get :$ $u_{k + 1} = \frac{3 + \frac{8}{2 k + 1}}{1 + \frac{2}{2 k + 1}} = \frac{6 k + 11}{2 k + 3} = 3 + \frac{2}{2 (k + 1) + 1}$ $This matches the formula for n = k + 1 .$ $Therefore, the solution to the recurrence relation is :$ $u_{n} = 3 + \frac{2}{2 n + 1}$
	Commented by mr W last updated on 13/Dec/24
	�� thanks!
	Commented by mr W last updated on 13/Dec/24
	$i′ll post a more general method with which we can solve questions like { ((u_(n+1) =((au_n +b)/(u_n +c)))),((u_0 =d)) :}$
	$i^{'} l l p o s t a m o r e g e n e r a l m e t h o d w i t h$ $w h i c h w e c a n s o l v e q u e s t i o n s l i k e$ ${\begin{cases} u_{n + 1} = \frac{{a u}_{n} + b}{u_{n} + c} \\ u_{0} = d \end{cases}$
	Answered by mr W last updated on 13/Dec/24

	$u_{n + 1} = \frac{4 (u_{n} - 2) - 1}{u_{n} - 2} = 4 - \frac{1}{u_{n} - 2}$ $u_{n + 1} - 2 = 2 - \frac{1}{u_{n} - 2}$ $l e t u_{n} - 2 = \frac{a_{n}}{b_{n}}, t h e n$ $\frac{a_{n + 1}}{b_{n + 1}} = 2 - \frac{b_{n}}{a_{n}} = \frac{2 a_{n} - b_{n}}{a_{n}}$ $\Rightarrow b_{n + 1} = a_{n}$ $\Rightarrow a_{n + 1} = 2 a_{n} - b_{n} = 2 a_{n} - a_{n - 1}$ $\Rightarrow a_{n + 1} - 2 a_{n} + a_{n - 1} = 0$ $c h a r a c t e r i s t i c e q u a t i o n$ $r^{2} - 2 r + 1 = 0 \Rightarrow r_{1, 2} = 1$ $\Rightarrow a_{n} = A + n B$ $b_{n} = a_{n - 1} = A + (n - 1) B$ $u_{n} - 2 = \frac{A + n B}{A + (n - 1) B} = \frac{C + n}{C + n - 1}$ $\Rightarrow u_{n} = 2 + \frac{C + n}{C + n - 1} = 3 + \frac{1}{C + n - 1}$ $u_{0} = 3 + \frac{1}{C - 1} = 5 \Rightarrow C = \frac{3}{2}$ $\Rightarrow u_{n} = 3 + \frac{1}{\frac{3}{2} + n - 1} = 3 + \frac{2}{2 n + 1} ✓$
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