Resolve-u-n-3u-n-1-12-3-4-n-and-u-n-2u-n-1-5cos-n-3-u-o-1-

Question Number 160529 by LEKOUMA last updated on 01/Dec/21

R e s o l v e

u_{n} - 3 u_{n - 1} = 12 {(\frac{3}{4})}^{n} a n d

u_{n} = 2 u_{n - 1} + 5 \cos (n \frac{Π}{3}), u_{o} = 1

Answered by mr W last updated on 02/Dec/21

(2)

u_{n} = 2 u_{n - 1} + 5 \cos \frac{n π}{3}

l e t u_{n} = v_{n} + A \sin \frac{n π}{3}

v_{n} + A \sin \frac{n π}{3} = 2 v_{n - 1} + 2 A \sin \frac{(n - 1) π}{3} + 5 \cos \frac{n π}{3}

v_{n} = 2 v_{n - 1} + 2 A \sin \frac{(n - 1) π}{3} - A \sin \frac{n π}{3} + 5 \cos \frac{n π}{3}

l e t 2 A \sin \frac{(n - 1) π}{3} - A \sin \frac{n π}{3} + 5 \cos \frac{n π}{3} = 0

A (\sin \frac{n π}{3} - \sqrt{3} \cos \frac{n π}{3}) - A \sin \frac{n π}{3} + 5 \cos \frac{n π}{3} = 0

(5 - A \sqrt{3}) \cos \frac{n π}{3} = 0

\Rightarrow A = \frac{5}{\sqrt{3}}

\Rightarrow u_{n} = v_{n} + \frac{5}{\sqrt{3}} \sin \frac{n π}{3}

v_{n} = 2 v_{n - 1} = 2^{2} v_{n - 2} = \dots = 2^{n} v_{0}

u_{0} = v_{0} + \frac{5}{\sqrt{3}} \sin \frac{0 π}{3} = 1 \Rightarrow v_{0} = 1

\Rightarrow v_{n} = 2^{n}

\Rightarrow u_{n} = 2^{n} + \frac{5}{\sqrt{3}} \sin \frac{n π}{3}

Answered by mr W last updated on 01/Dec/21

(1)

u_{n} - 3 u_{n - 1} = 12 {(\frac{3}{4})}^{n}

l e t u_{n} = v_{n} + k {(\frac{3}{4})}^{n}

v_{n} + k {(\frac{3}{4})}^{n} - 3 v_{n - 1} - 3 k {(\frac{3}{4})}^{n - 1} = 12 {(\frac{3}{4})}^{n}

v_{n} - 3 v_{n - 1} - 3 (k + 4) {(\frac{3}{4})}^{n} = 0

l e t k = - 4, i . e . u_{n} = v_{n} - 4 {(\frac{3}{4})}^{n}, t h e n

v_{n} - 3 v_{n - 1} = 0

v_{n} = 3 v_{n - 1} = 3^{2} v_{n - 2} = \dots = 3^{n} v_{0}

u_{0} = v_{0} - 4 {(\frac{3}{4})}^{0} = 1 \Rightarrow v_{0} = 5

\Rightarrow v_{n} = 5 \times 3^{n}

\Rightarrow u_{n} = 5 \times 3^{n} - 4 {(\frac{3}{4})}^{n} = 3^{n} (5 - \frac{1}{4^{n - 1}})

Commented by Tawa11 last updated on 01/Dec/21

Great sir .

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