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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

	$l e t a_{n} + {p b}_{n} = q (a_{n - 1} + {p b}_{n - 1})$ $a_{n} + {p b}_{n} = (3 + p) a_{n - 1} + 2 (1 + p) b_{n - 1}$ $\Rightarrow 3 + p = q$ $\Rightarrow 2 (1 + p) = p q$ $2 (1 + p) = p (3 + p)$ $p^{2} + p - 2 = 0$ $(p - 1) (p + 2) = 0$ $\Rightarrow p = 1, q = 4$ $\Rightarrow p = - 2, q = 1$ $a_{n} + b_{n} = 4 (a_{n - 1} + b_{n - 1}) = 4^{n} (a_{0} + b_{0}) = 3 \times 4^{n}$ $a_{n} = 3 a_{n - 1} + 2 (3 \times 4^{n - 1} - a_{n - 1})$ $a_{n} = a_{n - 1} + 6 \times 4^{n - 1}$ $a_{n - 1} = a_{n - 2} + 6 \times 4^{n - 2}$ $. . .$ $a_{1} = a_{0} + 6 \times 4^{0}$ $\Rightarrow a_{n} = a_{0} + 6 (4^{0} + 4^{1} + 4^{2} + . . . + 4^{n - 1})$ $\Rightarrow a_{n} = 1 + 6 \times \frac{4^{n} - 1}{4 - 1}$ $\Rightarrow a_{n} = 2 \times 4^{n} - 1$ $\Rightarrow a_{n} = 2^{2 n + 1} - 1$ $b_{n} = 3 \times 4^{n} - (2^{2 n + 1} - 1)$ $\Rightarrow b_{n} = 2^{2 n} + 1$
	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
	${ ((a_n =3a_(n−1) +2b_(n−1) ((a_n ),(b_n ) ))),((b_n =a_(n−1) +2b_(n−1) ⇒)) :}= (((3 2)),((1 2)) ) ((a_(n−1) ),(b_(n−1) ) ) ⇒ ((a_n ),(b_n ) ) =M^n ((a_0 ),(b_0 ) ) with M = (((3 2)),((1 2)) ) the problem lead to find M^n p_c (x)=det(M−xI)= determinant (((3−x 2)),((1 2−x)))=(3−x)(2−x)−2 =6−3x−2x+x^2 −2 =x^2 −5x+4 =x^2 −1 −5x+5 =(x−1)(x+1)−5(x−1) =(x−1)(x−4) so propre values are λ_1 =1 and λ_2 =4 let divide x^(n ) by P_c (M) ⇒ x^n =P_c (x)q(x)+u_n x +v_n ⇒1^n =u_n +v_n and4^n =4u_n +v_n ⇒ { ((u_n +v_n =1 { ((u_n =((4^n −1)/3))),((v_n =1−((4^n −1)/3))) :})),((4u_n +v_n =4^(n ) ⇒)) :} ⇒ { ((u_n =((4^n −1)/3) ⇒ { ((u_n =((4^n −1)/3))),((v_n =((4−4^n )/3))) :})),((v_n =((3−4^n +1)/3))) :} M^n =u_n M+v_n I_2 (cayley P_c (M)=0) ⇒M^n =((4^n −1)/3) (((3 2)),((1 2)) ) +((4−4^n )/3) (((1 0)),((0 1)) ) = (((4^n −1 (2/3)(4^n −1))),((((4^n −1)/3) (2/3)(4^n −1))) ) + (((((4−4^n )/3) 0)),((0 ((4−4^n )/3))) ) = (((((3.4^n −3+4−4^n )/3) (2/3)(4^n −1))),((((4^n −1)/3) ((2.4^n −2+4−4^n )/3))) ) = (((((2.4^n +1)/3) (2/3)(4^n −1))),((((4^n −1)/3) ((4^n +2)/3))) ) ⇒ ((a_n ),(b_n ) )=(1/3) (((2.4^n +1 2.4^n −2)),((4^n −1 4^n +2)) ) . ((1),(2) ) =(1/3) (((2.4^n +1+4.4^n −4)),((4^n −1 +2.4^n +4)) )=(1/3) (((6.4^n −3)),((3.4^n +3)) ) = (((2.4^n −1)),((4^n +1)) ) ⇒ a_n =2.4^n −1 and v_n =4^n +1$
	${\begin{cases} a_{n} = {3 a}_{n - 1} + {2 b}_{n - 1} (\begin{array}{c} a_{n} \\ b_{n} \end{array}) \\ b_{n} = a_{n - 1} + {2 b}_{n - 1} \Rightarrow \end{cases} = (\begin{matrix} 3 2 \\ 1 2 \end{matrix}) (\begin{matrix} a_{n - 1} \\ b_{n - 1} \end{matrix})$ $\Rightarrow (\begin{matrix} a_{n} \\ b_{n} \end{matrix}) = M^{n} (\begin{matrix} a_{0} \\ b_{0} \end{matrix}) with M = (\begin{matrix} 3 2 \\ 1 2 \end{matrix}) the problem lead to find M^{n}$ $p_{c} (x) = \det (M - xI) = \| \begin{matrix} 3 - x 2 \\ 1 2 - x \end{matrix} \| = (3 - x) (2 - x) - 2$ $= 6 - 3 x - 2 x + x^{2} - 2 = x^{2} - 5 x + 4 = x^{2} - 1 - 5 x + 5$ $= (x - 1) (x + 1) - 5 (x - 1) = (x - 1) (x - 4) so propre values are$ $λ_{1} = 1 and λ_{2} = 4 let divide x^{n} by P_{c} (M) \Rightarrow$ $x^{n} = P_{c} (x) q (x) + u_{n} x + v_{n} \Rightarrow 1^{n} = u_{n} + v_{n} {and 4}^{n} = {4 u}_{n} + v_{n} \Rightarrow$ ${\begin{cases} u_{n} + v_{n} = 1 {\begin{cases} u_{n} = \frac{4^{n} - 1}{3} \\ v_{n} = 1 - \frac{4^{n} - 1}{3} \end{cases} \\ {4 u}_{n} + v_{n} = 4^{n} \Rightarrow \end{cases}$ $\Rightarrow {\begin{cases} u_{n} = \frac{4^{n} - 1}{3} \Rightarrow {\begin{cases} u_{n} = \frac{4^{n} - 1}{3} \\ v_{n} = \frac{4 - 4^{n}}{3} \end{cases} \\ v_{n} = \frac{3 - 4^{n} + 1}{3} \end{cases}$ $M^{n} = u_{n} M + v_{n} I_{2} (cayley P_{c} (M) = 0)$ $\Rightarrow M^{n} = \frac{4^{n} - 1}{3} (\begin{matrix} 3 2 \\ 1 2 \end{matrix}) + \frac{4 - 4^{n}}{3} (\begin{matrix} 1 0 \\ 0 1 \end{matrix})$ $= (\begin{matrix} 4^{n} - 1 \frac{2}{3} (4^{n} - 1) \\ \frac{4^{n} - 1}{3} \frac{2}{3} (4^{n} - 1) \end{matrix}) + (\begin{matrix} \frac{4 - 4^{n}}{3} 0 \\ 0 \frac{4 - 4^{n}}{3} \end{matrix})$ $= (\begin{matrix} \frac{3 . 4^{n} - 3 + 4 - 4^{n}}{3} \frac{2}{3} (4^{n} - 1) \\ \frac{4^{n} - 1}{3} \frac{2 . 4^{n} - 2 + 4 - 4^{n}}{3} \end{matrix})$ $= (\begin{matrix} \frac{2 . 4^{n} + 1}{3} \frac{2}{3} (4^{n} - 1) \\ \frac{4^{n} - 1}{3} \frac{4^{n} + 2}{3} \end{matrix})$ $\Rightarrow (\begin{matrix} a_{n} \\ b_{n} \end{matrix}) = \frac{1}{3} (\begin{matrix} 2 . 4^{n} + 1 2 . 4^{n} - 2 \\ 4^{n} - 1 4^{n} + 2 \end{matrix}) . (\begin{matrix} 1 \\ 2 \end{matrix})$ $= \frac{1}{3} (\begin{matrix} 2 . 4^{n} + 1 + 4 . 4^{n} - 4 \\ 4^{n} - 1 + 2 . 4^{n} + 4 \end{matrix}) = \frac{1}{3} (\begin{matrix} 6 . 4^{n} - 3 \\ 3 . 4^{n} + 3 \end{matrix})$ $= (\begin{matrix} 2 . 4^{n} - 1 \\ 4^{n} + 1 \end{matrix}) \Rightarrow a_{n} = 2 . 4^{n} - 1 and v_{n} = 4^{n} + 1$
	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

	$Thanks sir, i appreciate$
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