if u_n =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n ] then u_(n+1) =u_n +u


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Question Number 194579 by MM42 last updated on 10/Jul/23

$i f u_{n} = \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}]$ $t h e n u_{n + 1} = u_{n} + u_{n - 1} ?; n = 0, 1, 2, . .$
	Answered by Frix last updated on 10/Jul/23
	$n={0, 1, 2, 3, 4, 5, 6, 7, ...} u_n ={0, 1, 1, 2, 3, 5, 8, 13, ...}$
	$n = {0, 1, 2, 3, 4, 5, 6, 7, . . .}$ $u_{n} = {0, 1, 1, 2, 3, 5, 8, 13, . . .}$
	Commented by MM42 last updated on 10/Jul/23

	$f i b o n a c c i s e q u e n c e$ $p l e a s e p r o v e . . .$
	Answered by MM42 last updated on 13/Jul/23

	$p r o v e$ $u_{n} + u_{n - 1} = \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n} + {(\frac{1 + \sqrt{5}}{2})}^{n - 1} - {(\frac{1 - \sqrt{5}}{2})}^{n - 1}]$ $= \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{n - 1} (\frac{1 + \sqrt{5}}{2} + 1) - {(\frac{1 - \sqrt{5}}{2})}^{n - 1} (\frac{1 - \sqrt{5}}{2} + 1)]$ $= \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{n - 1} (\frac{3 + \sqrt{5}}{2}) - {(\frac{1 - \sqrt{5}}{2})}^{n - 1} (\frac{3 - \sqrt{5}}{2})]$ $= \frac{1}{\sqrt{5}} [{(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n}] = u_{n + 1}$ $m e t o d 2$ $l e t a = \frac{1 + \sqrt{5}}{2} & b = \frac{1 - \sqrt{5}}{2} \Rightarrow u_{n} = \frac{1}{\sqrt{5}} (a^{n} - b^{n})$ $a + b = 1 & a b = - 1 \Rightarrow x^{2} + x - 1 = 0; a, b, a r e t h e r o o t s$ $a + b = 1 \Rightarrow a^{n + 1} + a^{n} b = a^{n} \Rightarrow a^{n + 1} - a^{n - 1} = a^{n} (i)$ $b^{n + 1} + {a b}^{n} = b^{n} \Rightarrow b^{n + 1} - b^{n - 1} = b^{n} (i i)$ $(i) + (i i) \Rightarrow a^{n + 1} - b^{n + 1} = (a^{n} - b^{n}) + (a^{n - 1} - b^{n - 1})$ $\Rightarrow u_{n + 1} = u_{m} + u_{n - 1}$
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