Given a_(n+1) = 3−6a_n and a_4 =−9 Find a


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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

	$a_{n + 1} + 6 a_{n} = 3$ $l e t a_{n} = b_{n} + k$ $b_{n + 1} + k + 6 (b_{n} + k) = 3$ $b_{n + 1} + 6 b_{n} = 3 - 7 k \overset{!}{=} 0 \Rightarrow k = \frac{3}{7}$ $b_{n} = - 6 b_{n - 1} = {(- 6)}^{2} b_{n - 2} = . . . = {(- 6)}^{n - 4} b_{4}$ $a_{n} - k = {(- 6)}^{n - 4} (a_{4} - k)$ $a_{n} = {(- 6)}^{n - 4} (a_{4} - k) + k$ $a_{n} = {(- 6)}^{n - 4} (- 9 - \frac{3}{7}) + \frac{3}{7}$ $\Rightarrow a_{n} = \frac{11 {(- 6)}^{n - 3} + 3}{7}$
	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
	$⇒a_(n+1) +6a_n −3=0 he→r+6=0 ⇒r=−6 ⇒a_n =λ(−6)^n +ρ a_4 =λ(−6)^4 +ρ a_5 =3−6a_4 =3−6(−9)=3+6.9 =57=λ(−6)^5 +ρ we get the system { (((−6)^4 λ+ρ =−9)),(((−6)^5 λ +ρ =57 ⇒)) :} ((−6)^5 −(−6)^4 )λ=57+9 =66 ⇒ λ=((66)/(−6^5 −6^4 ))=−((66)/(6^4 +6^5 ))=−((66)/(6^4 ×7)) ρ=−9−6^4 λ =−9+6^4 .((66)/(6^4 .7)) =−9+((66)/7) rest to finich to calculus...$
	$\Rightarrow a_{n + 1} + 6 a_{n} - 3 = 0$ $h e \to r + 6 = 0 \Rightarrow r = - 6 \Rightarrow a_{n} = λ {(- 6)}^{n} + ρ$ $a_{4} = λ {(- 6)}^{4} + ρ$ $a_{5} = 3 - 6 a_{4} = 3 - 6 (- 9) = 3 + 6.9$ $= 57 = λ {(- 6)}^{5} + ρ w e g e t t h e s y s t e m$ ${\begin{cases} {(- 6)}^{4} λ + ρ = - 9 \\ {(- 6)}^{5} λ + ρ = 57 \Rightarrow \end{cases}$ $({(- 6)}^{5} - {(- 6)}^{4}) λ = 57 + 9 = 66 \Rightarrow$ $λ = \frac{66}{- 6^{5} - 6^{4}} = - \frac{66}{6^{4} + 6^{5}} = - \frac{66}{6^{4} \times 7}$ $ρ = - 9 - 6^{4} λ = - 9 + 6^{4} . \frac{66}{6^{4} . 7}$ $= - 9 + \frac{66}{7} r e s t t o f i n i c h t o c a l c u l u s . . .$
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