Given a sequence (u_n ) defined reculsively by u_(n+1) = 3u_n + 4u_(n−1) , u_0 = 1 , u_2 = 3 show that u_(n+1) −4u_n = (−1)^(n+1) (3u_0 −4u_1 ) hence show that u


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Question Number 127116 by physicstutes last updated on 27/Dec/20

$Given a sequence (u_{n}) defined reculsively by$ $u_{n + 1} = 3 u_{n} + 4 u_{n - 1}, u_{0} = 1, u_{2} = 3$ $show that u_{n + 1} - 4 u_{n} = {(- 1)}^{n + 1} (3 u_{0} - 4 u_{1})$ $hence show that u_{n} is a divegent sequence .$
Commented by 676597498 last updated on 27/Dec/20

$f . m a t h s g c e 2020 .$ $i g o t t h e m a r k g u i d e$
Commented by Ar Brandon last updated on 27/Dec/20
Quel pays ?
	Answered by mr W last updated on 27/Dec/20

	$p^{2} - 3 p - 4 = 0$ $(p - 4) (p + 1) = 0$ $p = 4, - 1$ $u_{n} = A \times 4^{n} + B \times {(- 1)}^{n}$ $u_{2} = 16 A + B = 3$ $u_{0} = A + B = 1$ $\Rightarrow A = \frac{2}{15}$ $\Rightarrow B = \frac{13}{15}$ $u_{n} = \frac{2}{15} \times 4^{n} + \frac{13}{15} {(- 1)}^{n}$ $u_{1} = \frac{8}{15} - \frac{13}{15} = - \frac{1}{3}$ $3 u_{0} - 4 u_{1} = 3 + \frac{4}{3} = \frac{13}{3} = 5 B$ $u_{n + 1} - 4 u_{n} = A \times 4^{n + 1} + B \times {(- 1)}^{n + 1} - 4 A \times 4^{n} - 4 B \times {(- 1)}^{n}$ $= 5 B \times {(- 1)}^{n + 1} = {(- 1)}^{n + 1} (3 u_{0} - 4 u_{1})$
	Commented by physicstutes last updated on 27/Dec/20

	$thats wonderful gracias$
	Answered by Raxreedoroid last updated on 27/Dec/20

	$Given a sequence (u_{n}) defined reculsively by$ $u_{n + 1} = 3 u_{n} + 4 u_{n - 1}, u_{0} = 1, u_{2} = 3$ $show that u_{n + 1} - 4 u_{n} = {(- 1)}^{n + 1} (3 u_{0} - 4 u_{1})$ $hence show that u_{n} is a divegent sequence .$ $3 u_{1} = u_{2} - 4 u_{0}$ $u_{1} = \frac{3 - 4}{3} = - 0 . \overline{3}$ $u_{n + 1} - 4 u_{n} = {(- 1)}^{n + 1} (3 \times 1 + 4 \times 0 . \overline{3})$ $u_{n + 1} - 4 u_{n} = 4 . \bar{3} {(- 1)}^{n + 1}$ $L e t n = 1 t h e n u_{2} - 4 u_{1} = {(- 1)}^{1 + 1} (4 . \overline{3})$ $3 - 4 (- 0 . \bar{3}) = 4 . \bar{3}$ $4 . \bar{3} = 4 . \bar{3} ✓$ $(a) : A s s u m e u_{n + 1} - 4 u_{n} = 4 . \bar{3} {(- 1)}^{n + 1}$ $S u b t i t u t e x a s 4 . \bar{3} {(- 1)}^{n + 1} u_{n + 1} - 4 u_{n} = x$ $s o l v e f o r x$ $u_{n + 1} = 3 u_{n} + 4 u_{n - 1} ∴ 3 u_{n} + 4 u_{n - 1} - 4 u_{n} = x$ $x = 4 u_{n - 1} - u_{n}$ $∴ - x = u_{n} - 4 u_{n - 1}$ $u_{n} - 4 u_{n - 1} = 4 . \bar{3} (- 1) {(- 1)}^{n + 1}$ $u_{n} - 4 u_{n - 1} = 4 . \bar{3} {(- 1)}^{n + 2} = 4 . \bar{3} {(- 1)}^{2} {(- 1)}^{n} = 4 . \bar{3} {(- 1)}^{n}$ $S u b s i t u t e n a s n + 1$ $u_{n + 1} - 4 u_{n} = 4 . \bar{3} {(- 1)}^{n + 1} w h i c h i s t r u e f r o m (a)$ $u_{n + 1} = 4 . \bar{3} {(- 1)}^{n + 1} + 4 u_{n}$ $= 4 . \bar{3} {(- 1)}^{n + 1} + 4 \times 4 . \bar{3} {(- 1)}^{n} + 16 u_{n - 1} . . . 4^{n} {(- 1)}^{n + 1} + 4^{n + 1}$ $u_{n + 1} = 4^{n + 1} + \underset{i = 1}{\sum^{n + 1}} 4^{i - 1} \times 4 . \bar{3} {(- 1)}^{n - i}$ $u_{n} = 4^{n} + 4 . \bar{3} \underset{i = 1}{\sum^{n}} 4^{i - 1} {(- 1)}^{n - i - 1}$ $u_{n} = 4^{n} + 4 . \bar{3} \underset{i = 1}{\sum^{n}} 4^{i - 1} {(- 1)}^{n - i - 1}$ $\underset{n \to \infty}{l i m} (4^{n} + 4 . \bar{3} \underset{i = 1}{\sum^{n}} 4^{i - 1} {(- 1)}^{n - i - 1}) = \infty$
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