Given-a-n-1-3-6a-n-and-a-4-9-Find-a-n-

Question Number 206997 by efronzo1 last updated on 03/May/24

Given a_{n + 1} = 3 - {6 a}_{n} and a_{4} = - 9

Find a_{n} = ?

Answered by mr W last updated on 04/May/24

a_{n + 1} + 6 a_{n} = 3

l e t a_{n} = b_{n} + k

b_{n + 1} + k + 6 (b_{n} + k) = 3

b_{n + 1} + 6 b_{n} = 3 - 7 k \overset{!}{=} 0 \Rightarrow k = \frac{3}{7}

b_{n} = - 6 b_{n - 1} = {(- 6)}^{2} b_{n - 2} = \dots = {(- 6)}^{n - 4} b_{4}

a_{n} - k = {(- 6)}^{n - 4} (a_{4} - k)

a_{n} = {(- 6)}^{n - 4} (a_{4} - k) + k

a_{n} = {(- 6)}^{n - 4} (- 9 - \frac{3}{7}) + \frac{3}{7}

\Rightarrow a_{n} = \frac{11 {(- 6)}^{n - 3} + 3}{7}

Commented by efronzo1 last updated on 03/May/24

\Rightarrow a_{n + 1} = b_{n + 1} + k

\Rightarrow 3 - a_{n} = b_{n + 1} + k

\Rightarrow 3 - (b_{n} + k) = b_{n + 1} + k

\Rightarrow b_{n + 1} + b_{n} = 3 - 2 k

Commented by mr W last updated on 03/May/24

b u t y o u r q u e s t i o n i s

\Rightarrow 3 - {6 a}_{n} = b_{n + 1} + k

Commented by efronzo1 last updated on 03/May/24

why 3 - 7 k = 0 ?

Commented by mr W last updated on 03/May/24

w e j u s t s e t 3 - 7 k = 0 s u c h t h a t

b_{n} b e c o m e s a n e a s y G . P . :

b_{n} = - 6 b_{n - 1}

Commented by mathzup last updated on 03/May/24

t h e m e t h o d o f s i r m r w i s c o r r e c t

Answered by mathzup last updated on 03/May/24

\Rightarrow a_{n + 1} + 6 a_{n} - 3 = 0

h e \to r + 6 = 0 \Rightarrow r = - 6 \Rightarrow a_{n} = λ {(- 6)}^{n} + ρ

a_{4} = λ {(- 6)}^{4} + ρ

a_{5} = 3 - 6 a_{4} = 3 - 6 (- 9) = 3 + 6.9

= 57 = λ {(- 6)}^{5} + ρ w e g e t t h e s y s t e m

{\begin{cases} {(- 6)}^{4} λ + ρ = - 9 \\ {(- 6)}^{5} λ + ρ = 57 \Rightarrow \end{cases}

({(- 6)}^{5} - {(- 6)}^{4}) λ = 57 + 9 = 66 \Rightarrow

λ = \frac{66}{- 6^{5} - 6^{4}} = - \frac{66}{6^{4} + 6^{5}} = - \frac{66}{6^{4} \times 7}

ρ = - 9 - 6^{4} λ = - 9 + 6^{4} . \frac{66}{6^{4} . 7}

= - 9 + \frac{66}{7} r e s t t o f i n i c h t o c a l c u l u s \dots

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