Find-the-solution-of-recurrence-relation-a-n-2a-n-1-3a-n-2-with-a-0-1-a-1-2-

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

\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 \dots (i)

\Rightarrow a_{1} = - A + 3 B = 2 \dots (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

Facebook Tweet Pin