if-a-n-2-a-n-1-3-a-n-2-and-a-1-2-a-2-4-find-a-n-

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}

\dots .

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 \dots