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Question Number 81911 by jagoll last updated on 16/Feb/20

Commented by jagoll last updated on 16/Feb/20

$a_{n + 1} = \frac{3 a_{n} + 1}{2}$ $a_{n + 1} + A = A + \frac{3 a_{n} + 1}{2} = \frac{3 a_{n} + 2 A + 1}{2}$ $a_{n + 1} + B = B + \frac{3 a_{n} + 1}{2} = \frac{3 a_{n} + 2 B + 1}{2}$ $\frac{a_{n + 1} + A}{a_{n + 1} + B} = \frac{3 a_{n} + 2 A + 1}{3 a_{n} + 2 B + 1}$ $\frac{a_{n + 1} + A}{a_{n + 1} + B} = \frac{3}{3} \times \frac{a_{n} + \frac{2 A + 1}{3}}{a_{n} + \frac{2 B + 1}{3}}$ $A = \frac{2 A + 1}{3} \Rightarrow A = 1$ $B = 1 . i h a v e t r o u b l e$
Commented by jagoll last updated on 16/Feb/20

$h a h a h a . . w h a t w r o n g t h i s m y w o r k ?$
Commented by mr W last updated on 16/Feb/20

$t h e t a c t i c w i t h a_{n + 1} + A a n d a_{n + 1} + B$ $i s d u e t o t h e f a c t t h a t w e h a v e a_{n} i n$ $a_{n + 1} = \frac{. . . a_{n} + . . . .}{. . . a_{n} + . . .}$ $y o u c a n n o t a p p l y t h i s t a c t i c f o r o t h e r$ $c a s e s .$
Commented by jagoll last updated on 17/Feb/20

$o k a y s i r t h a n k y o u$
	Answered by mr W last updated on 16/Feb/20

	$a_{n + 1} = \frac{3}{2} a_{n} + \frac{1}{2}$ $a_{n + 1} + A = \frac{3}{2} a_{n} + \frac{1}{2} + A$ $a_{n + 1} + A = \frac{3}{2} (a_{n} + \frac{1}{3} + \frac{2 A}{3})$ $\Rightarrow A \overset{!}{=} \frac{1}{3} + \frac{2 A}{3}$ $\Rightarrow A = 1$ $\Rightarrow a_{n + 1} + 1 = \frac{3}{2} (a_{n} + 1)$ $l e t b_{n} = a_{n} + 1$ $\Rightarrow b_{n + 1} = \frac{3}{2} b_{n} \leftarrow {i t}^{'} s a G P$ $b_{1} = a_{1} + 1 = 2 + 1 = 3$ $\Rightarrow b_{n} = b_{1} {(\frac{3}{2})}^{n - 1} = 3 {(\frac{3}{2})}^{n - 1} = \frac{3^{n}}{2^{n - 1}} = a_{n} + 1$ $\Rightarrow a_{n} = \frac{3^{n}}{2^{n - 1}} - 1$
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