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-n-is-a-divegent-sequence-

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} \dots 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