if a_1 =2 a_(n+1) =a_n ^2 −1 find a


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Question Number 80645 by mr W last updated on 05/Feb/20

$i f$ $a_{1} = 2$ $a_{n + 1} = a_{n}^{2} - 1$ $f i n d a_{n} = ?$
Commented by jagoll last updated on 05/Feb/20

$a_{2} = 3, a_{3} = 8, a_{4} = 63$ $a_{5} = 63^{2} - 1$
Commented by mr W last updated on 05/Feb/20

$i t i s t o f i n d a f o r m u l a f o r a_{n} i n$ $t e r m s o f n .$
Commented by jagoll last updated on 05/Feb/20

$i h a v e n o t y e t g o t a n i d e a t o$ $f i n i s h$
Commented by jagoll last updated on 05/Feb/20

$h o w t o s o l v e i t s i r ?$
Commented by mind is power last updated on 05/Feb/20

$i w i l l t r y$
Commented by mr W last updated on 06/Feb/20

$t h i s q u e s t i o n i s t o o h a r d, i h a v e n o i d e a$ $h o w t o s o l v e .$ $b u t i t w e r e m u c h e a s i e r, i f t h e q u e s t i o n$ $w e r e a_{n + 1} = a_{n}^{2} - 2 .$
Commented by jagoll last updated on 06/Feb/20

$w h a t t h e a n s w e r s i h i f ?$ $a_{n + 1} = a_{n}^{2} - 2$
Commented by mr W last updated on 06/Feb/20

$s a y a_{1} = 3, a_{n + 1} = a_{n}^{2} - 2$ $a_{n} = p^{2^{n}} + \frac{1}{p^{2 n}}$ $a_{n}^{2} = {(p^{2^{n}} + \frac{1}{p^{2 n}})}^{2} = p^{2^{n + 1}} + 2 + \frac{1}{p^{2^{n + 1}}}$ $a_{n}^{2} - 2 = p^{2^{n + 1}} + \frac{1}{p^{2^{n + 1}}} = a_{n + 1}$ $a_{1} = p^{2} + \frac{1}{p^{2}} = 3$ ${(p - \frac{1}{p})}^{2} = 1$ $p - \frac{1}{p} = \pm 1$ $p^{2} \pm p - 1 = 0$ $p = \frac{\pm 1 \pm \sqrt{5}}{2}$ $\Rightarrow a_{n} = {(\frac{1 + \sqrt{5}}{2})}^{2^{n}} + {(\frac{1 - \sqrt{5}}{2})}^{2^{n}}$
Commented by jagoll last updated on 08/Feb/20

$h o w t o g e t a_{n} = p^{2 n} + \frac{1}{p^{2 n}} s i r$
Commented by mr W last updated on 08/Feb/20

$n o t a_{n} = p^{2 n} + \frac{1}{p^{2 n}}, b u t a_{n} = p^{2^{n}} + \frac{1}{p^{2^{n}}} .$ $i n t u i t i v e w e k n o w t h e t e r m s m u s t b e$ $i n t h e f o r m a_{1} = k + \frac{1}{k}, s i n c e w e h a v e$ $a_{1}^{2} - 2 = k^{2} + 2 + \frac{1}{k^{2}} - 2 = k^{2} + \frac{1}{k^{2}} = a_{2} . a n d$ $a_{2}^{2} - 2 = k^{4} + 2 + \frac{1}{k^{4}} - 2 = k^{4} + \frac{1}{k^{4}} = a_{3} . t h e$ $r e s t i s c l e a r, a_{n} = k^{2^{n - 1}} + \frac{1}{k^{2^{n - 1}}} = p^{2^{n}} + \frac{1}{p^{2^{n}}} .$
Commented by mr W last updated on 08/Feb/20

${t h a t}^{'} s w h y i s a i d i t i s o n l y p o s s i b l e f o r$ $a_{n + 1} = a_{n}^{2} - 2 . y o u c a n n o t a p p l y t h i s f o r$ $a_{n + 1} = a_{n}^{2} - 1 .$
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