Question-128849

Question Number 128849 by I want to learn more last updated on 10/Jan/21

Commented by Tawa11 last updated on 15/Sep/21

nice

Answered by mr W last updated on 10/Jan/21

l e t a_{n} + {p b}_{n} = q (a_{n - 1} + {p b}_{n - 1})

a_{n} + {p b}_{n} = (3 + p) a_{n - 1} + 2 (1 + p) b_{n - 1}

\Rightarrow 3 + p = q

\Rightarrow 2 (1 + p) = p q

2 (1 + p) = p (3 + p)

p^{2} + p - 2 = 0

(p - 1) (p + 2) = 0

\Rightarrow p = 1, q = 4

\Rightarrow p = - 2, q = 1

a_{n} + b_{n} = 4 (a_{n - 1} + b_{n - 1}) = 4^{n} (a_{0} + b_{0}) = 3 \times 4^{n}

a_{n} = 3 a_{n - 1} + 2 (3 \times 4^{n - 1} - a_{n - 1})

a_{n} = a_{n - 1} + 6 \times 4^{n - 1}

a_{n - 1} = a_{n - 2} + 6 \times 4^{n - 2}

\dots

a_{1} = a_{0} + 6 \times 4^{0}

\Rightarrow a_{n} = a_{0} + 6 (4^{0} + 4^{1} + 4^{2} + \dots + 4^{n - 1})

\Rightarrow a_{n} = 1 + 6 \times \frac{4^{n} - 1}{4 - 1}

\Rightarrow a_{n} = 2 \times 4^{n} - 1

\Rightarrow a_{n} = 2^{2 n + 1} - 1

b_{n} = 3 \times 4^{n} - (2^{2 n + 1} - 1)

\Rightarrow b_{n} = 2^{2 n} + 1

Commented by I want to learn more last updated on 10/Jan/21

Thanks sir, i appreciate .

Answered by mathmax by abdo last updated on 11/Jan/21

{\begin{cases} a_{n} = {3 a}_{n - 1} + {2 b}_{n - 1} (\begin{array}{c} a_{n} \\ b_{n} \end{array}) \\ b_{n} = a_{n - 1} + {2 b}_{n - 1} \Rightarrow \end{cases} = (\begin{matrix} 3 2 \\ 1 2 \end{matrix}) (\begin{matrix} a_{n - 1} \\ b_{n - 1} \end{matrix})

\Rightarrow (\begin{matrix} a_{n} \\ b_{n} \end{matrix}) = M^{n} (\begin{matrix} a_{0} \\ b_{0} \end{matrix}) with M = (\begin{matrix} 3 2 \\ 1 2 \end{matrix}) the problem lead to find M^{n}

p_{c} (x) = \det (M - xI) = | \begin{matrix} 3 - x 2 \\ 1 2 - x \end{matrix} | = (3 - x) (2 - x) - 2

= 6 - 3 x - 2 x + x^{2} - 2 = x^{2} - 5 x + 4 = x^{2} - 1 - 5 x + 5

= (x - 1) (x + 1) - 5 (x - 1) = (x - 1) (x - 4) so propre values are

λ_{1} = 1 and λ_{2} = 4 let divide x^{n} by P_{c} (M) \Rightarrow

x^{n} = P_{c} (x) q (x) + u_{n} x + v_{n} \Rightarrow 1^{n} = u_{n} + v_{n} {and 4}^{n} = {4 u}_{n} + v_{n} \Rightarrow

{\begin{cases} u_{n} + v_{n} = 1 {\begin{cases} u_{n} = \frac{4^{n} - 1}{3} \\ v_{n} = 1 - \frac{4^{n} - 1}{3} \end{cases} \\ {4 u}_{n} + v_{n} = 4^{n} \Rightarrow \end{cases}

\Rightarrow {\begin{cases} u_{n} = \frac{4^{n} - 1}{3} \Rightarrow {\begin{cases} u_{n} = \frac{4^{n} - 1}{3} \\ v_{n} = \frac{4 - 4^{n}}{3} \end{cases} \\ v_{n} = \frac{3 - 4^{n} + 1}{3} \end{cases}

M^{n} = u_{n} M + v_{n} I_{2} (cayley P_{c} (M) = 0)

\Rightarrow M^{n} = \frac{4^{n} - 1}{3} (\begin{matrix} 3 2 \\ 1 2 \end{matrix}) + \frac{4 - 4^{n}}{3} (\begin{matrix} 1 0 \\ 0 1 \end{matrix})

= (\begin{matrix} 4^{n} - 1 \frac{2}{3} (4^{n} - 1) \\ \frac{4^{n} - 1}{3} \frac{2}{3} (4^{n} - 1) \end{matrix}) + (\begin{matrix} \frac{4 - 4^{n}}{3} 0 \\ 0 \frac{4 - 4^{n}}{3} \end{matrix})

= (\begin{matrix} \frac{3 . 4^{n} - 3 + 4 - 4^{n}}{3} \frac{2}{3} (4^{n} - 1) \\ \frac{4^{n} - 1}{3} \frac{2 . 4^{n} - 2 + 4 - 4^{n}}{3} \end{matrix})

= (\begin{matrix} \frac{2 . 4^{n} + 1}{3} \frac{2}{3} (4^{n} - 1) \\ \frac{4^{n} - 1}{3} \frac{4^{n} + 2}{3} \end{matrix})

\Rightarrow (\begin{matrix} a_{n} \\ b_{n} \end{matrix}) = \frac{1}{3} (\begin{matrix} 2 . 4^{n} + 1 2 . 4^{n} - 2 \\ 4^{n} - 1 4^{n} + 2 \end{matrix}) . (\begin{matrix} 1 \\ 2 \end{matrix})

= \frac{1}{3} (\begin{matrix} 2 . 4^{n} + 1 + 4 . 4^{n} - 4 \\ 4^{n} - 1 + 2 . 4^{n} + 4 \end{matrix}) = \frac{1}{3} (\begin{matrix} 6 . 4^{n} - 3 \\ 3 . 4^{n} + 3 \end{matrix})

= (\begin{matrix} 2 . 4^{n} - 1 \\ 4^{n} + 1 \end{matrix}) \Rightarrow a_{n} = 2 . 4^{n} - 1 and v_{n} = 4^{n} + 1

Commented by I want to learn more last updated on 11/Jan/21

Thanks sir, I appreciate .

Commented by I want to learn more last updated on 11/Jan/21

Thanks sir, i appreciate

Answered by Olaf last updated on 11/Jan/21

X_{0} = (\begin{matrix} 1 \\ 2 \end{matrix}), A = [\begin{matrix} 3 & 2 \\ 1 & 2 \end{matrix}], X_{n} = {AX}_{n - 1}

X_{n} = A^{n} X_{0} = (\begin{matrix} a_{n} \\ b_{n} \end{matrix})

We solve \det (A - λ I) = 0

\Rightarrow (3 - λ) (2 - λ) - 2 = 0

eigenvalues :

λ_{1} = 1 and λ_{2} = 4

B = [\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 4 \end{matrix}]

A (\begin{matrix} x \\ y \end{matrix}) = λ_{1} (\begin{matrix} x \\ y \end{matrix}) \Rightarrow x_{1} = - 1 and y_{1} = 1

A (\begin{matrix} x \\ y \end{matrix}) = λ_{2} (\begin{matrix} x \\ y \end{matrix}) \Rightarrow x_{2} = 2 and y_{2} = 1

eigenvectors :

V_{1} = (\begin{matrix} - 1 \\ 1 \end{matrix}) and V_{2} = (\begin{matrix} 2 \\ 1 \end{matrix})

P = [\begin{matrix} - 1 & 2 \\ 1 & 1 \end{matrix}]

P^{- 1} = [\begin{matrix} - \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} \end{matrix}]

B = P^{- 1} AP and A^{n} = {PB}^{n} P^{- 1}

A^{n} = [\begin{matrix} - 1 & 2 \\ 1 & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & 4^{n} \end{matrix}] [\begin{matrix} - \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} \end{matrix}]

A^{n} = [\begin{matrix} - 1 & 2 \\ 1 & 1 \end{matrix}] [\begin{matrix} - \frac{1}{3} & \frac{2}{3} \\ \frac{4^{n}}{3} & \frac{4^{n}}{3} \end{matrix}]

A^{n} = [\begin{matrix} \frac{2 . 4^{n} + 1}{3} & \frac{2 . 4^{n} - 2}{3} \\ \frac{4^{n} - 1}{3} & \frac{4^{n} + 2}{3} \end{matrix}]

A^{n} = [\begin{matrix} \frac{2^{2 n + 1} + 1}{3} & \frac{2^{2 n + 1} - 2}{3} \\ \frac{2^{2 n} - 1}{3} & \frac{2^{2 n} + 2}{3} \end{matrix}]

X_{n} = A^{n} X_{0} = (\begin{matrix} \frac{2^{2 n + 1} + 1}{3} + 2 \frac{2^{2 n + 1} - 2}{3} \\ \frac{2^{2 n} - 1}{3} + 2 \frac{2^{2 n} + 2}{3} \end{matrix})

X_{n} = A^{n} X_{0} = (\begin{matrix} 2^{2 n + 1} - 1 \\ 2^{2 n} + 1 \end{matrix})

Commented by I want to learn more last updated on 11/Jan/21

Thanks sir, i appreciate