if a_(n+2) =(a_(n+1) ^3 /a_n ^2 ) and a_1 =2, a_2 =4 find a


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Question Number 55869 by mr W last updated on 05/Mar/19

$i f a_{n + 2} = \frac{a_{n + 1}^{3}}{a_{n}^{2}} a n d a_{1} = 2, a_{2} = 4$ $f i n d a_{n} = ?$
Commented by Tawa1 last updated on 05/Mar/19

$sir, please check question 55785 you solved . I asked something . .$
Commented by MJS last updated on 05/Mar/19

$a_{n} = \frac{a_{2}^{(2^{n - 1} - 1)}}{a_{1}^{(2^{n - 1} - 2)}}$
	Answered by MJS last updated on 05/Mar/19

	$a_{n} = 2^{(2^{n - 1})} = 2^{\frac{2^{n}}{2}}$
	Commented by mr W last updated on 05/Mar/19

	$t h a n k s s i r!$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

	$\frac{a_{n + 2}}{a_{n + 1}} = {(\frac{a_{n + 1}}{a_{n}})}^{2}$ $\frac{a_{3}}{a_{2}} = {(\frac{a_{2}}{a_{1}})}^{2} = {(\frac{4}{2})}^{2} = 4$ $\frac{a_{2}}{a_{1}} = 2 = 2^{1}$ $\frac{a_{3}}{a_{2}} = 4 = 2^{2}$ $\frac{a_{4}}{a_{3}} = {(\frac{a_{3}}{a_{2}})}^{2} = 4^{2} = 2^{4}$ $\frac{a_{5}}{a_{4}} = {(\frac{a_{4}}{a_{3}})}^{2} = 2^{8}$ $. . . .$ $a_{1} = a_{1} = 2 \to 2^{2^{0}}$ $a_{2} = 2^{1} a_{1} = 4 = 2^{2} \to 2^{2^{1}}$ $a_{3} = 2^{2} a_{2} = 2^{2} \times 2^{2} = 2^{4} \to 2^{2^{2}}$ $a_{4} = 2^{4} a_{3} = 2^{4} \times 2^{4} = 2^{8} \to 2^{2^{3}}$ $a_{5} = 2^{8} \times a^{4} = 2^{8} \times 2^{8} \to 2^{2^{4}}$ $s o a_{n} = 2^{2^{n - 1}}$
	Commented by mr W last updated on 05/Mar/19

	$v e r y n i c e s i r! t h a n k y o u!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19

	$t h a n k y o u s i r b r a i n c e l l g e t s c h a r g e d a n d g e t s a c t i v a t e d$ $t o s o l v e t h e s e t y p e o f p r o b l e m s . . .$
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