Given the series a_n = 3a_(n−1) + 2a_(n−2) with a_1 =11 & a_2 = 21. Find a


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Question Number 137899 by bemath last updated on 08/Apr/21

$G i v e n t h e s e r i e s$ $a_{n} = 3 a_{n - 1} + 2 a_{n - 2} w i t h$ $a_{1} = 11 & a_{2} = 21 . F i n d a_{n} .$
	Answered by benjo_mathlover last updated on 08/Apr/21

	$c h a r a c t e r i s t i c s e q u a t i o n$ $r^{2} - 3 r - 2 = 0; r = \frac{3 \pm \sqrt{17}}{2}$ $a_{n} = A . {(\frac{3 + \sqrt{17}}{2})}^{n} + B {(\frac{3 - \sqrt{17}}{2})}^{n}$ $w i t h n = 1 \Rightarrow 11 = \frac{3}{2} (A + B) + \frac{\sqrt{17}}{2} (A - B)$ $w i t h n = 2 \Rightarrow 21 = \frac{13}{2} (A + B) + \frac{3 \sqrt{17}}{2} (A - B)$ $A = (\frac{20 - 3 \sqrt{17}}{\sqrt{17}}); B = - (\frac{20 + 3 \sqrt{17}}{\sqrt{17}})$ $∴ a_{n} = (\frac{20 - 3 \sqrt{17}}{\sqrt{17}}) {(\frac{3 + \sqrt{17}}{2})}^{n} - (\frac{20 + 3 \sqrt{17}}{\sqrt{17}}) {(\frac{3 - \sqrt{17}}{2})}^{n}$
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