a_1 =4 a_(n+1) =((4a_n +3)/(a_n +2)) find a


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Question Number 81871 by mr W last updated on 16/Feb/20

$a_{1} = 4$ $a_{n + 1} = \frac{4 a_{n} + 3}{a_{n} + 2}$ $f i n d a_{n} = ?$
Commented by jagoll last updated on 16/Feb/20

$m y a n s w e r i s c o r r e c t m i s t e r ?$
Commented by john santu last updated on 16/Feb/20

$n o t c o r r e c t!$ $a_{n + 1} = 4 - \frac{5}{a_{n} + 2}$ $n = 1 \Rightarrow a_{2} = 4 - \frac{5}{6} = \frac{19}{6}$ $n = 2 \Rightarrow a_{3} = 4 - \frac{5}{(\frac{19}{6}) + 2} = 4 - \frac{30}{31}$ $= \frac{124 - 30}{31} = \frac{94}{31} \neq \frac{43}{72} (27) - \frac{53}{24}$
Commented by jagoll last updated on 16/Feb/20

$h a h a . . . w r o n g s i r$
	Answered by mr W last updated on 16/Feb/20

	$a_{n + 1} + A = \frac{4 a_{n} + 3}{a_{n} + 2} + A = \frac{(4 + A) a_{n} + (3 + 2 A)}{a_{n} + 2}$ $a_{n + 1} + B = \frac{(4 + B) a_{n} + (3 + 2 B)}{a_{n} + 2}$ $\frac{a_{n + 1} + A}{a_{n + 1} + B} = \frac{(4 + A) a_{n} + (3 + 2 A)}{(4 + B) a_{n} + (3 + 2 B)}$ $\frac{a_{n + 1} + A}{a_{n + 1} + B} = \frac{4 + A}{4 + B} \times \frac{a_{n} + \frac{3 + 2 A}{4 + A}}{a_{n} + \frac{3 + 2 B}{4 + B}}$ $A = \frac{3 + 2 A}{4 + A} \Rightarrow A^{2} + 2 A - 3 = 0$ $\Rightarrow (A + 3) (A - 1) = 0 \Rightarrow A = - 3 o r 1$ $s i m i l a r l y$ $\Rightarrow B = - 3 o r 1$ $w e t a k e A \neq B, e . g . A = - 3, B = 1,$ $\frac{a_{n + 1} - 3}{a_{n + 1} + 1} = \frac{1}{5} (\frac{a_{n} - 3}{a_{n} + 1})$ $l e t b_{n} = \frac{a_{n} - 3}{a_{n} + 1}$ $\Rightarrow b_{n + 1} = \frac{1}{5} b_{n} \leftarrow t h i s i s a G . P .!$ $\Rightarrow b_{n} = b_{1} {(\frac{1}{5})}^{n - 1}$ $b_{1} = \frac{a_{1} - 3}{a_{1} + 1} = \frac{4 - 3}{4 + 1} = \frac{1}{5}$ $\Rightarrow b_{n} = {(\frac{1}{5})}^{n} = \frac{a_{n} - 3}{a_{n} + 1}$ $\Rightarrow a_{n} = \frac{3 + {(\frac{1}{5})}^{n}}{1 - {(\frac{1}{5})}^{n}} = \frac{3 \times 5^{n} + 1}{5^{n} - 1}$
	Commented by jagoll last updated on 16/Feb/20

	$a l r i g h t, i^{'} l l t r y t o u s e t h i s m e t h o d$ $f o r s i m i l a r p r o b l e m s i n m y b o o k,$ $m i s t e r$
	Commented by jagoll last updated on 16/Feb/20

	$w h y i s i t d i f f e r e n t f r o m t h e$ $p r e v i o u s p r o b l e m ? t h a t y o u p o s t$ $t o o . w h a t i s t h e g e n e r a l$ $f o r m o f s o l v i n g t h i s t y p e o f$ $p r o b l e m ?$
	Commented by mr W last updated on 16/Feb/20

	$f o r t y p e a_{n + 1} = 2 a_{n} - 3 a_{n - 1} y o u c a n$ $s e e t h a t t h e t e r m s h o u l d b e i n f o r m$ $a_{n} = {A q}^{n}, s i n c e i t f u l f u l l s t h e e q n .$ $a_{n + 1} = 2 a_{n} - 3 a_{n - 1} . b u t y o u c a n n o t a p p l y$ $i t f o r o t h e r t y p e s l i k e a_{n + 1} = (n + 1) a_{n} - 2 a_{n - 1} .$ $t h a t m e a n s t h e r e a r e n o u n i v e r s a l$ $m e t h o d s! o n e s h o u l d t r y d i f f e r e n t$ $m e t h o d s f r o m c a s e t o c a s e . f o r$ $s a m e t y p e s y o u c a n f i n d n o s o l u t i o n,$ $e . g . a_{n + 1} = 2 a_{n}^{2} + 1, a_{n + 1} = \sqrt{a_{n} + 2} + 1 e t c .$
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