if a_0 =1, a_1 =2 and a_(n+1) =(√(a_n a_(n−1) )) find a


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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

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