if-a-0-1-a-1-2-and-a-n-1-a-n-a-n-1-find-a-n-in-terms-of-n-

Question Number 130593 by mr W last updated on 27/Jan/21

i f a_{0} = 1, a_{1} = 2 a n d a_{n + 1} = \sqrt{a_{n} a_{n - 1}}

f i n d a_{n} i n t e r m s o f n .

Commented by malwan last updated on 27/Jan/21

a_{n} = 2^{(\frac{(\frac{2^{n} + {(- 1)}^{n - 1}}{3})}{2^{n - 1}})}

n ⩾ 3 e d i t e d

Commented by malwan last updated on 27/Jan/21

a_{2} = {(2 \times 1)}^{\frac{1}{2}} = 2^{\frac{1}{2}}

a_{3} = {(2^{\frac{1}{2}} \times 2)}^{\frac{1}{2}} = 2^{\frac{3}{4}}

a_{4} = {(2^{\frac{3}{4}} \times 2^{\frac{1}{2}})}^{\frac{1}{2}} = 2^{\frac{5}{8}}

a_{5} = {(2^{\frac{5}{8}} \times 2^{\frac{3}{4}})}^{\frac{1}{2}} = 2^{\frac{11}{16}} = 2^{\frac{(\frac{32 + 1}{3})}{2^{4}}}

= 2^{\frac{(\frac{2^{5} + 1^{5 - 1}}{3})}{2^{5 - 1}}}

a_{6} = {(2^{\frac{11}{16}} \times 2^{\frac{5}{8}})}^{\frac{1}{2}} = 2^{\frac{21}{32}} = 2^{\frac{(\frac{64 - 1}{3})}{2^{5}}}

= 2^{\frac{(\frac{2^{6} + {(- 1)}^{6 - 1}}{3})}{2^{6 - 1}}}

a_{7} = {(2^{\frac{21}{32}} \times 2^{\frac{11}{16}})}^{\frac{1}{2}} = 2^{\frac{43}{64}} = 2^{\frac{(\frac{2^{7} + {(- 1)}^{7 - 1}}{3})}{2^{7 - 1}}}

∴ a_{n} = 2^{\frac{(\frac{2^{n} + {(- 1)}^{n - 1}}{3})}{2^{n - 1}}}

Commented by mr W last updated on 27/Jan/21

t h a n k y o u! g o o d!

Answered by mindispower last updated on 27/Jan/21

\Rightarrow l n (a_{n + 1}) - \frac{1}{2} l n (a_{n}) - \frac{1}{z} l n (a_{n - 1}) = 0

l n (a_{n}) . w_{n}

\Rightarrow w_{n + 1} - \frac{w_{n}}{2} - \frac{w_{n - 1}}{2} = 0

X^{2} - \frac{X}{2} - \frac{1}{2} . 0

X_{1} = 1, X_{2} = - \frac{1}{2}

w_{n} = a + b {(- \frac{1}{2})}^{n}

w_{0} = 0 \Rightarrow a + b = 0, 2 a - b = 2 l n (2)

a = \frac{2}{3} l n (2), b = - \frac{2}{3} l n (2)

w_{n} = l n (2^{\frac{2}{3}}) - \frac{2}{3} l n (2) {(- \frac{1}{2})}^{n}

= l n (2^{\frac{2}{3}} . 2^{\frac{- 2}{3} . \frac{{(- 1)}^{n}}{2^{n}}})

= l n (2^{\frac{2^{n + 1} - 2 {(- 1)}^{n}}{3 . 2^{n}}}) = w_{n}, a_{n} = e^{w_{n}} = 2^{\frac{2^{n + 1} - 2 {(- 1)}^{n}}{3 . 2^{n}}}

Commented by mr W last updated on 27/Jan/21

t h a n k y o u s i r!

{t h a t}^{'} s w h a t i a l s o g o t :

a_{n} = 2^{\frac{2}{3} (1 - \frac{{(- 1)}^{n}}{2^{n}})}

Commented by mindispower last updated on 27/Jan/21

a l w a y s p l e a s u r s i r