given-a-n-1-a-n-1-2a-n-if-a-3-0-amp-a-5-1-find-a-n-

Question Number 104428 by bemath last updated on 21/Jul/20

g i v e n a_{n + 1} = a_{n - 1} + 2 a_{n}

i f a_{3} = 0 & a_{5} = - 1

f i n d a_{n}

Answered by Dwaipayan Shikari last updated on 21/Jul/20

Put n = 4

a_{5} = a_{3} + {2 a}_{4}

- 1 = {2 a}_{4}

a_{4} = - \frac{1}{2}

Commented by bemath last updated on 21/Jul/20

a_{n} ?

Commented by bemath last updated on 21/Jul/20

a n y o n e h e l p m e ?

Answered by mathmax by abdo last updated on 22/Jul/20

a_{n + 1} = a_{n - 1} + {2 a}_{n} \Rightarrow a_{n + 2} = a_{n} + {2 a}_{n + 1} \Rightarrow a_{n + 2} - {2 a}_{n + 1} - a_{n} = 0

caracteristic equation r^{2} - 2 r - 1 = 0 \Rightarrow r^{2} - 2 r + 1 - 2 = 0 \Rightarrow

{(r - 1)}^{2} - {(\sqrt{2})}^{2} = 0 \Rightarrow (r - 1 - \sqrt{2}) (r - 1 + \sqrt{2}) = 0 \Rightarrow r = 1 + \sqrt{2} or r = 1 - \sqrt{2}

a_{n} = α {(1 + \sqrt{2})}^{n} + β {(1 - \sqrt{2})}^{n}

a_{3} = α {(1 + \sqrt{2})}^{3} + β {(1 - \sqrt{2})}^{3} = 0 and a_{5} = α {(1 + \sqrt{2})}^{5} + β {(1 - \sqrt{2})}^{5} = - 1 \Rightarrow

we get the system {\begin{cases} {(1 + \sqrt{2})}^{3} α + {(1 - \sqrt{2})}^{3} β = 0 \\ {(1 + \sqrt{2})}^{5} α + {(1 - \sqrt{2})}^{5} β = - 1 \end{cases}

Δ_{s} = {(1 + \sqrt{2})}^{3} {(1 - \sqrt{2})}^{5} - {(1 + \sqrt{2})}^{5} {(1 - \sqrt{2})}^{3}

= {(1 + \sqrt{2})}^{3} {(1 - \sqrt{2})}^{3} {(1 - \sqrt{2})}^{2} - {(1 + \sqrt{2})}^{2} {(1 - \sqrt{2})}^{3} {(1 + \sqrt{2})}^{3}

= - {(1 - \sqrt{2})}^{2} + {(1 + \sqrt{2})}^{2} = 3 + 2 \sqrt{2} - (3 - 2 \sqrt{2}) = 4 \sqrt{2} \neq 0 \Rightarrow

α = \frac{Δ_{α}}{Δ s} = \frac{| \begin{matrix} o {(1 - \sqrt{2})}^{3} \\ - 1 {(1 - \sqrt{2})}^{5} \end{matrix} |}{4 \sqrt{2}} = \frac{{(1 - \sqrt{2})}^{3}}{4 \sqrt{2}}

β = \frac{Δ_{β}}{Δ_{s}} = \frac{| \begin{matrix} {(1 + \sqrt{2})}^{3} 0 \\ {(1 + \sqrt{2})}^{5} - 1 \end{matrix} |}{4 \sqrt{2}} = - \frac{{(1 + \sqrt{2})}^{3}}{4 \sqrt{2}}

\Rightarrow a_{n} = \frac{{(1 - \sqrt{2})}^{3}}{4 \sqrt{2}} {(1 + \sqrt{2})}^{n} - \frac{{(1 + \sqrt{2})}^{3}}{4 \sqrt{2}} {(1 - \sqrt{2})}^{n}

Commented by bemath last updated on 22/Jul/20

t h a n k y o u s i r

Answered by john santu last updated on 21/Jul/20

Commented by bemath last updated on 21/Jul/20

w o w \dots

Commented by bemath last updated on 21/Jul/20

h o w t o g e t A a n d B s i r ?

Answered by mr W last updated on 21/Jul/20

a_{5} = 2 a_{4} + a_{3}

- 1 = 2 a_{4} + 0 \Rightarrow a_{4} = - \frac{1}{2}

a_{4} = 2 a_{3} + a_{2}

- \frac{1}{2} = 2 \times 0 + a_{2} \Rightarrow a_{2} = - \frac{1}{2}

a_{3} = 2 a_{2} + a_{1}

0 = 2 (- \frac{1}{2}) + a_{1} \Rightarrow a_{1} = 1

a_{2} = 2 a_{1} + a_{0}

- \frac{1}{2} = 2 \times 1 + a_{0} \Rightarrow a_{0} = - \frac{5}{2}

a_{n + 1} - 2 a_{n} - a_{n - 1} = 0

l e t a_{n} = C λ^{n}

C λ^{n + 1} - 2 C λ^{n} - C λ^{n - 1} = 0

C λ^{n - 1} (λ^{2} - 2 λ - 1) = 0

\Rightarrow λ^{2} - 2 λ - 1 = 0

\Rightarrow λ_{1} = 1 + \sqrt{2}

\Rightarrow λ_{2} = 1 - \sqrt{2}

\Rightarrow a_{n} = C_{1} {(1 + \sqrt{2})}^{n} + C_{2} {(1 - \sqrt{2})}^{n}

a_{0} = C_{1} + C_{2} = - \frac{5}{2}

a_{1} = C_{1} (1 + \sqrt{2}) + C_{2} (1 - \sqrt{2}) = 1

\Rightarrow C_{1} = \frac{7 \sqrt{2} - 10}{8}

\Rightarrow C_{2} = - \frac{7 \sqrt{2} + 10}{8}

\Rightarrow a_{n} = \frac{(7 \sqrt{2} - 10) {(1 + \sqrt{2})}^{n} - (7 \sqrt{2} + 10) {(1 - \sqrt{2})}^{n}}{8}

Commented by bemath last updated on 22/Jul/20

t h a n k y o u s i r

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