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Question Number 34355 by srikarkailash9@gmail.com last updated on 05/May/18

Commented by math khazana by abdo last updated on 05/May/18

$l e t p r o v e b y r e c u r r e n c e t h a t$ $A^{n} = (\begin{matrix} p^{n} q \frac{p^{n} - 1}{p - 1} \\ 0 1 \end{matrix})$ $w i t h p \neq 1 f o r n = 1 A^{1} = (\begin{matrix} p q \\ 0 1 \end{matrix})$ $r e l a t i o n t r u e l e t s u p p o s e A^{n} = (\begin{matrix} p^{n} q \frac{p^{n} - 1}{p - 1} \\ 0 1 \end{matrix})$ $A^{n + 1} = A^{n} . A = (\begin{matrix} p^{n} q \frac{p^{n} - 1}{p - 1} \\ 0 1 \end{matrix}) . (\begin{matrix} p q \\ 0 1 \end{matrix})$ $= (\begin{matrix} p^{n + 1} p^{n} q + q \frac{p^{n} - 1}{p - 1} \\ 0 1 \end{matrix})$ $= (\begin{matrix} p^{n + 1} q (\frac{p^{n + 1} - p^{n} + p^{n} - 1}{p - 1}) \\ 0 1 \end{matrix})$ $= (\begin{matrix} p^{n + 1} q \frac{p^{n + 1} - 1}{p - 1} \\ 0 1 \end{matrix})$ $t h e r e l a t i o n i s t r u e a t t e r m n + 1 l e t t a k e n = 8 \Rightarrow$ $A^{8} = (\begin{matrix} p^{8} q \frac{p^{8} - 1}{p - 1} \\ 0 1 \end{matrix})$
	Answered by srikarkailash9@gmail.com last updated on 05/May/18

	$s o l v e i t s t e p w i s e$
	Answered by candre last updated on 05/May/18
	$A^1 = ((p,q),(0,1) ) A^n = ((a_n ,b_n ),(c_n ,d_n ) ) A^(n+1) = ((a_(n+1) ,b_(n+1) ),(c_(n+1) ,d_(n+1) ) ) =A^n A = ((a_n ,b_n ),(c_n ,d_n ) ) ((p,q),(0,1) ) = (((a_n p),(a_n q+b_n )),((c_n p),(c_n q+d_n )) ) from where we get { ((a_(n+1) =a_n p),(a_1 =p)),((b_(n+1) =a_n q+b_n ),(b_1 =q)),((c_(n+1) =c_n p),(c_1 =0)),((d_(n+1) =c_n q+d_n ),(d_1 =0)) :} studing each separated for a a_2 =a_1 p a_3 =a_2 p=a_1 p^2 ⋮ a_n =a_1 p^(n−1) ⇒a_(n+1) =a_n p=a_1 p^n a_1 =p⇒a_n =pp^(n−1) =p^n for c similiar argument apply, but: c_1 =0⇒c_n =c_1 p^(n−1) =0 for b b_2 =a_1 q+b_1 b_3 =a_2 q+b_2 =(a_1 +a_2 )q+b_1 ⋮ b_n =(a_1 +...+a_(n−1) )q+b_1 ⇒b_(n+1) =a_n q+b_n =(a_1 +...+a_(n−1) +a_n )q+b_1 a_n =p^n ⇒Σ_(i=1) ^n a_i =p+p^2 +...+p^n =p((p^n −1)/(p−1))=((p^(n+1) −p)/(p−1))(p≠1) b_1 =q⇒b_n =qΣ_(i=1) ^(n−1) a_i +q=q(((p^n −p)/(p−1))+1)=q((p^n −p+p−1)/(p−1))=q((p^n −1)/(p−1)) similiar for d c_n =0⇒Σ_(i=1) ^n c_i =0 d_1 =1⇒d_n =qΣ_(i=1) ^(n−1) c_n +d_1 =1 so A^n = ((p^n ,(q((p^n −1)/(p−1)))),(0,1) ) where p≠1 above is just the case of n=8$
	$A^{1} = (\begin{matrix} p & q \\ 0 & 1 \end{matrix})$ $A^{n} = (\begin{matrix} a_{n} & b_{n} \\ c_{n} & d_{n} \end{matrix})$ $A^{n + 1} = (\begin{matrix} a_{n + 1} & b_{n + 1} \\ c_{n + 1} & d_{n + 1} \end{matrix})$ $= A^{n} A$ $= (\begin{matrix} a_{n} & b_{n} \\ c_{n} & d_{n} \end{matrix}) (\begin{matrix} p & q \\ 0 & 1 \end{matrix})$ $= (\begin{matrix} a_{n} p & a_{n} q + b_{n} \\ c_{n} p & c_{n} q + d_{n} \end{matrix})$ $from where we get$ ${\begin{cases} a_{n + 1} = a_{n} p & a_{1} = p \\ b_{n + 1} = a_{n} q + b_{n} & b_{1} = q \\ c_{n + 1} = c_{n} p & c_{1} = 0 \\ d_{n + 1} = c_{n} q + d_{n} & d_{1} = 0 \end{cases}$ $studing each separated$ $for a$ $a_{2} = a_{1} p$ $a_{3} = a_{2} p = a_{1} p^{2}$ $⋮$ $a_{n} = a_{1} p^{n - 1} \Rightarrow a_{n + 1} = a_{n} p = a_{1} p^{n}$ $a_{1} = p \Rightarrow a_{n} = {p p}^{n - 1} = p^{n}$ $for c similiar argument apply, b u t :$ $c_{1} = 0 \Rightarrow c_{n} = c_{1} p^{n - 1} = 0$ $for b$ $b_{2} = a_{1} q + b_{1}$ $b_{3} = a_{2} q + b_{2} = (a_{1} + a_{2}) q + b_{1}$ $⋮$ $b_{n} = (a_{1} + . . . + a_{n - 1}) q + b_{1} \Rightarrow b_{n + 1} = a_{n} q + b_{n} = (a_{1} + . . . + a_{n - 1} + a_{n}) q + b_{1}$ $a_{n} = p^{n} \Rightarrow \underset{i = 1}{\sum^{n}} a_{i} = p + p^{2} + . . . + p^{n} = p \frac{p^{n} - 1}{p - 1} = \frac{p^{n + 1} - p}{p - 1} (p \neq 1)$ $b_{1} = q \Rightarrow b_{n} = q \underset{i = 1}{\sum^{n - 1}} a_{i} + q = q (\frac{p^{n} - p}{p - 1} + 1) = q \frac{p^{n} - p + p - 1}{p - 1} = q \frac{p^{n} - 1}{p - 1}$ $similiar for d$ $c_{n} = 0 \Rightarrow \underset{i = 1}{\sum^{n}} c_{i} = 0$ $d_{1} = 1 \Rightarrow d_{n} = q \underset{i = 1}{\sum^{n - 1}} c_{n} + d_{1} = 1$ $s o$ $A^{n} = (\begin{matrix} p^{n} & q \frac{p^{n} - 1}{p - 1} \\ 0 & 1 \end{matrix}) where p \neq 1$ $above is just the case of n = 8$
	Commented by candre last updated on 05/May/18

	$if p = 1 we get$ $A^{n} = (\begin{matrix} 1 & q n \\ 0 & 1 \end{matrix})$ $;)$
	Commented by abdo mathsup 649 cc last updated on 05/May/18

	$w h e r e a r e y o u f r o m s i r c a n d r e ?$
	Commented by candre last updated on 05/May/18

	$b r a z i l, b u t w h y t h e q u e s t i o n ?$
	Commented by math khazana by abdo last updated on 05/May/18

	$i t s o n l y a s i m l p l e i n f o r m a t i o n s i r i k n o w t h a t$ $b r a z i l i s a b e a u t i f u l c o u n t r y w i t h g o o d a n d$ $g e n e r o u s p e o p l e i a m l o o k i n g t h a t y o u a r e m o s t$ $i n t e r e s t e d t o m a t h e m a t i c s . . . i h o p e t o v i s i t b r a z i l$ $s o m e d a y . . . .$
	Commented by Rio Mike last updated on 06/May/18

	$h e y i l i v e i n B r a z i l t o S a o P a o l o$ $c a n i k n o w y o u s i r ?$
	Commented by candre last updated on 11/May/18

	$i does live in state of sao paulo, but not the city .$
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