{ ((u_0 = 2)),((u_(n+1) = ((2u_n −1)/u_n ))) :} Find u


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Question Number 184048 by cortano1 last updated on 02/Jan/23
${ ((u_0 = 2)),((u_(n+1) = ((2u_n −1)/u_n ))) :} Find u_n .$
${\begin{cases} u_{0} = 2 \\ u_{n + 1} = \frac{2 u_{n} - 1}{u_{n}} \end{cases}$ $F i n d u_{n} .$
	Answered by SEKRET last updated on 02/Jan/23

	$n = 0 u_{1} = \frac{3}{2} n = 1 u_{2} = \frac{4}{3}$ $n = 2 u_{3} = \frac{5}{4} n = 3 u_{4} = \frac{6}{5}$ $n = 4 u_{5} = \frac{7}{6}$ $\frac{2}{1}; \frac{3}{2}; \frac{4}{3}; \frac{5}{4}; \frac{6}{5}; \frac{7}{6}$ $U_{n} = \frac{n + 2}{n + 1}$ $A B D U L A Z I Z A B D U V A L I Y E V$
	Commented by mr W last updated on 02/Jan/23

	$i s t h i s a s t r i c t m a t h e m a t i c a l p r o o f ?$ $i {d o n}^{'} t m e a n y o u r a n s w e r i s w r o n g,$ $b u t j u s t w a n t t o s a y t h a t o n e c o u l d$ $f a l l i n t o a t r a p u s i n g t h i s m e t h o d .$ $e x a m p l e : i n h o w m a n y r e g i o n s c a n a$ $c i r c l e a t m o s t b e d i v i d e d b y n p o i n t s$ $o n i t ? w e g e t$ $n = 1 : 1 r e g i o n$ $n = 2 : 2 r e g i o n s$ $n = 3 : 4 r e g i o n s$ $n = 4 : 8 r e g i o n s$ $n = 5 : 16 r e g i o n s$ $t h e n c a n w e s a y g e n e r a l l y$ $n p o i n t s \Rightarrow 2^{n - 1} r e g i o n s ?$
	Commented by Acem last updated on 02/Jan/23

	$i f u_{n} = \frac{n + 2}{n + 1} t h e n u_{0} = \frac{0 + 2}{0 + 1} = 2 ✓$ $i f u_{n} = \frac{n + 2}{n + 1} t h e n u_{n + 1} = \frac{n + 3}{n + 2}$ $u_{n + 1} = \frac{2 u_{n} - 1}{u_{n}} = \frac{2 \frac{n + 2}{n + 1} - 1}{\frac{n + 2}{n + 1}} = \frac{2 n + 4 - n - 1}{n + 2} = \frac{n + 3}{n + 2} ✓$
	Commented by mr W last updated on 02/Jan/23

	$n i c e s i r!$ $t h i s i s a s t r i c t p r o o f f o r u_{n} = \frac{n + 2}{n + 1}$ $w h e n w e a l r e a d y k n o w u_{n} = \frac{n + 2}{n + 1} .$ $b u t h o w t o f i n d u_{n} g e n e r a l l y ? {l e t}^{'} s$ $s a y u_{0} = c . c c o u l d b e a n y r e a l v a l u e .$
	Commented by SEKRET last updated on 02/Jan/23

	$thanks sir .$
	Commented by Acem last updated on 02/Jan/23

	$H e l l o S i r!$ $F i n d i n g u_{n} g e n e r a l l y, s h o u l d b e b y g u e s s i n g l a w$
	Commented by mr W last updated on 02/Jan/23

	$u_{n} = 1 + \frac{1}{n + \frac{1}{u_{0} - 1}}$ $s e e s o l u t i o n b e l o w .$
	Answered by cortano2 last updated on 02/Jan/23

	$u_{0} = 2$ $u_{1} = 3 / 2$ $u_{2} = 4 / 3$ $u_{3} = 5 / 4$ $u_{n} = \frac{n + 2}{n + 1}$
	Commented by mr W last updated on 02/Jan/23

	$i s t h i s a s t r i c t m a t h e m a t i c a l p r o o f ?$
	Commented by Acem last updated on 02/Jan/23

	$i f u_{n} = \frac{n + 2}{n + 1} t h e n u_{0} = \frac{0 + 2}{0 + 1} = 2 ✓$ $i f u_{n} = \frac{n + 2}{n + 1} t h e n u_{n + 1} = \frac{n + 3}{n + 2}$ $u_{n + 1} = \frac{2 u_{n} - 1}{u_{n}} = \frac{2 \frac{n + 2}{n + 1} - 1}{\frac{n + 2}{n + 1}} = \frac{2 n + 4 - n - 1}{n + 2} = \frac{n + 3}{n + 2} ✓$
	Answered by a.lgnaoui last updated on 02/Jan/23

	$u_{n} = 2 - \frac{1}{u_{n - 1}} = \frac{2 u_{n - 1} - 1}{u_{n - 1}}$ $u_{1} = \frac{4 - 1}{2} = \frac{3}{2} u_{2} = 2 - \frac{2}{3} = \frac{4}{3}$ $u_{3} = 2 - \frac{3}{4} = \frac{5}{4} u_{4} = 2 - \frac{4}{5} = \frac{6}{5}$ $u_{5} = 2 - \frac{5}{6} = \frac{7}{6} u_{6} = 2 - \frac{6}{7} = \frac{8}{7}$ $u_{7} = 2 - \frac{7}{8} = \frac{9}{8} u_{8} = 2 - \frac{8}{9} = \frac{10}{9}$ $. . . . . . . . . . .$ $d o n c u_{n} = 2 - \frac{n}{n + 1} = \frac{n + 2}{n + 1}$
	Answered by CrispyXYZ last updated on 02/Jan/23

	$u_{n} = \frac{n + 2}{n + 1}$ $Proof with Mathematical Induction$ $(1) n = 0$ $u_{0} = \frac{2}{1} = 2$ $(2) n = k \Rightarrow n = k + 1$ $u_{k} = \frac{k + 2}{k + 1}$ $u_{k + 1} = \frac{2 u_{k} - 1}{u_{k}} = 2 - \frac{k + 1}{k + 2} = \frac{k + 3}{k + 2} = \frac{(k + 1) + 2}{(k + 1) + 1}$ $(1) (2) \Rightarrow u_{n} = \frac{n + 2}{n + 1} is true for all n$
	Answered by mr W last updated on 02/Jan/23

	$u_{n + 1} = \frac{2 u_{n} - 1}{u_{n}}$ $u_{n + 1} - 1 = \frac{u_{n} - 1}{u_{n}}$ $\frac{1}{u_{n + 1} - 1} = \frac{u_{n}}{u_{n} - 1}$ $\frac{1}{u_{n + 1} - 1} = 1 + \frac{1}{u_{n} - 1}$ $\frac{1}{u_{n + 1} - 1} - \frac{1}{u_{n} - 1} = 1$ $l e t A_{n} = \frac{1}{u_{n} - 1}$ $A_{n + 1} - A_{n} = 1 \Rightarrow A . P .$ $A_{n} - A_{n - 1} = 1$ $. . .$ $A_{2} - A_{1} = 1$ $A_{1} - A_{0} = 1$ $\Rightarrow A_{n} - A_{0} = n$ $\Rightarrow A_{n} = A_{0} + n$ $\Rightarrow \frac{1}{u_{n} - 1} = \frac{1}{u_{0} - 1} + n$ $\Rightarrow u_{n} = 1 + \frac{1}{n + \frac{1}{u_{0} - 1}}$ $i n c a s e u_{0} = 2 :$ $u_{n} = 1 + \frac{1}{n + 1} = \frac{n + 2}{n + 1}$
	Commented by Acem last updated on 02/Jan/23

	$N i c e S i r! t h a n k s$ $q u e s t i o n, b e c m y E n g . {w h a t}^{'} s A . P .$
	Commented by cortano1 last updated on 03/Jan/23

	$a r i t m e t i c p r o g r e s s i o n$
	Commented by mr W last updated on 03/Jan/23

	$A . P . = a r i t h m e t h i c p r o g r e s s i o n$ $G . P . = g e o m e t r i c p r o g r e s s i o n$
	Commented by Acem last updated on 04/Jan/23

	$T h x f r i e n d s!$
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