Find the solution of recurrence relation a_n = 2a_(n−1) + 3a_(n−2) , with a_0 = 1, a


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Question Number 56864 by Joel578 last updated on 25/Mar/19

$Find the solution of recurrence relation$ $a_{n} = 2 a_{n - 1} + 3 a_{n - 2}, with a_{0} = 1, a_{1} = 2$
Commented by maxmathsup by imad last updated on 25/Mar/19
$⇒a_(n+2) =2a_(n+1) +3 a_n ⇒a_(n+2) −2a_(n+1) −3a_n =0 ⇒caracteristic equation is x^2 −2x−3 =0 ⇒Δ^′ =1+3=4 ⇒x_1 =1+2=3 and x_2 =1−2=−1 ⇒ a_n =α 3^n +β (−1)^n initial conditions a_0 =1=α+β and a_1 =2=3α −β ⇒α+β =1 and 3α−β=2 ⇒3α−(1−α)=2 ⇒ 4α =3 ⇒α=(3/4) ⇒β=1−(3/4) =(1/4) ⇒a_n =(3/4) 3^n +(((−1)^n )/4) ⇒ a_n =(1/4){3^(n+1) +(−1)^n }$
$\Rightarrow a_{n + 2} = 2 a_{n + 1} + 3 a_{n} \Rightarrow a_{n + 2} - 2 a_{n + 1} - 3 a_{n} = 0 \Rightarrow c a r a c t e r i s t i c e q u a t i o n i s$ $x^{2} - 2 x - 3 = 0 \Rightarrow Δ^{^{'}} = 1 + 3 = 4 \Rightarrow x_{1} = 1 + 2 = 3 a n d x_{2} = 1 - 2 = - 1 \Rightarrow$ $a_{n} = α 3^{n} + β {(- 1)}^{n} i n i t i a l c o n d i t i o n s$ $a_{0} = 1 = α + β a n d a_{1} = 2 = 3 α - β \Rightarrow α + β = 1 a n d 3 α - β = 2 \Rightarrow 3 α - (1 - α) = 2 \Rightarrow$ $4 α = 3 \Rightarrow α = \frac{3}{4} \Rightarrow β = 1 - \frac{3}{4} = \frac{1}{4} \Rightarrow a_{n} = \frac{3}{4} 3^{n} + \frac{{(- 1)}^{n}}{4} \Rightarrow$ $a_{n} = \frac{1}{4} {3^{n + 1} + {(- 1)}^{n}}$
Commented by Joel578 last updated on 26/Mar/19

$t h a n k y o u v e r y m u c h$
Commented by turbo msup by abdo last updated on 26/Mar/19

$y o u a r e w e l c o m e$
	Answered by mr W last updated on 25/Mar/19

	$l e t a_{n} = {A p}^{n} + {B q}^{n}$ $\Rightarrow {A p}^{n} + {B q}^{n} = 2 {A p}^{n - 1} + 2 {B q}^{n - 1} + 3 {A p}^{n - 2} + 3 {B q}^{n - 2}$ $\Rightarrow A (p^{2} - 2 p - 3) p^{n - 2} + B (q^{2} - 2 q - 3) q^{n - 2} = 0$ $\Rightarrow p a n d q a r e r o o t s o f$ $x^{2} - 2 x - 3 = 0$ $(x + 1) (x - 3) = 0$ $\Rightarrow p = - 1 a n d q = 3$ $\Rightarrow a_{n} = A \times {(- 1)}^{n} + B \times 3^{n}$ $\Rightarrow a_{0} = A + B = 1 . . . (i)$ $\Rightarrow a_{1} = - A + 3 B = 2 . . . (i i)$ $\Rightarrow A = \frac{1}{4}$ $\Rightarrow B = \frac{3}{4}$ $\Rightarrow a_{n} = \frac{1}{4} [{(- 1)}^{n} + 3^{n + 1}]$
	Commented by Joel578 last updated on 26/Mar/19

	$t h a n k y o u v e r y m u c h$
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