Given a_(n+1) = a_n + a_(n+2) where a_3 = 4 and a_5 = 6 find a


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Question Number 217190 by efronzo1 last updated on 05/Mar/25

$Given a_{n + 1} = a_{n} + a_{n + 2}$ $where a_{3} = 4 and a_{5} = 6$ $find a_{n} .$
	Answered by mr W last updated on 05/Mar/25

	$a_{n + 2} - a_{n + 1} + a_{n} = 0$ $c h a r a c t e r i s t i c e q u a t i o n :$ $r^{2} - r + 1 = 0$ $\Rightarrow r_{1, 2} = \frac{1 \pm i \sqrt{3}}{2} = e^{\pm \frac{π i}{3}}$ $s a y a_{n} = {A e}^{\frac{n π i}{3}} + {B e}^{- \frac{n π i}{3}}$ $o r a_{n} = C \cos \frac{n π}{3} + D \sin \frac{n π}{3}$ $a_{3} = - C = 4 \Rightarrow C = - 4$ $a_{5} = - 4 \cos \frac{5 π}{3} + D \sin \frac{5 π}{3} = 6 \Rightarrow D = - \frac{16}{\sqrt{3}}$ $\Rightarrow a_{n} = - 4 (\cos \frac{n π}{3} + \frac{4}{\sqrt{3}} \sin \frac{n π}{3})$ $o r$ $a_{n} = - \frac{4 \sqrt{57}}{3} \sin (\frac{n π}{3} + \tan^{- 1} \frac{\sqrt{3}}{4})$
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