if a_n −5a_(n−1) +6a_(n−2) =1 and a_0 =1, a_1 =2. find a


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Question Number 140102 by mr W last updated on 04/May/21

$i f a_{n} - 5 a_{n - 1} + 6 a_{n - 2} = 1 a n d$ $a_{0} = 1, a_{1} = 2 .$ $f i n d a_{n} i n t e r m s o f n .$
Commented by som(math1967) last updated on 04/May/21

$I s i t \frac{3^{n} + 1}{2} s i r ?$
Commented by mr W last updated on 04/May/21

$y e s, c o r r e c t .$
Commented by mr W last updated on 04/May/21

$n o t c o r r e c t s i r!$ $j u s t c h e c k a_{2}, a_{3} . t h e y s h o u l d b e :$ $a_{2} - 5 \times 2 + 6 \times 1 = 1 \Rightarrow a_{2} = 5$ $a_{3} - 5 \times 5 + 6 \times 2 = 1 \Rightarrow a_{3} = 14$
Commented by som(math1967) last updated on 04/May/21

$a_{n} - 5 a_{n - 1} + 6 a_{n - 2} = 1$ $n = 2$ $a_{2} = 1 + 5 \times 2 - 6 = 5$ $a_{3} = 14 . . . . a_{4} = 41$ $a_{0}, a_{1}, a_{2}, a_{3}$ $1, (1 + 3^{0}), (1 + 3^{0} + 3^{1}), (1 + 3^{0} + 3 + 3^{2})$ $. . . . \underset{a_{n}}{\underset{⏟}{(1 + 3^{0} + 3 + 3^{2} + . . . .)}}$ $a_{n} = 1 + \frac{3^{n} - 1}{3 - 1} = \frac{3^{n} + 1}{2}$
Commented by mr W last updated on 04/May/21

$t h i s m e t h o d i s n o t g e n e r a l e n o u g h .$ $i f i h a d g i v e n a_{1} = 1, a_{2} = 3, i t w o u l d$ $n o t b e s o e a s y t o g e t t h e s o l u t i o n u s i n g$ $t h i s m e t h o d .$ $a n y w a y, t h a n k s a l o t!$
Commented by Dwaipayan Shikari last updated on 04/May/21

$Σ a_{n} x^{n} - 5 Σ a_{n - 1} x^{n} + 6 Σ a_{n - 2} x^{n} = Σ x^{n}$ $\Rightarrow \underset{n = 0}{\sum^{\infty}} a_{n + 2} x^{n} - 5 \underset{n = 0}{\sum^{\infty}} a_{n + 1} x^{n} + 6 \underset{n = 0}{\sum^{\infty}} a_{n} x^{n} = Σ x^{n}$ $\Rightarrow (a_{2} x + a_{3} x^{2} + . .) - 5 (a_{1} x + a_{2} x^{2} + . . .) + 6 (a_{0} + a_{1} x + a_{2} x^{2} + . . .) = \frac{1}{1 - x}$ $\Rightarrow \frac{1}{x} (a_{0} + a_{1} x + a_{2} x^{2} + . . .) - \frac{a_{0}}{x} - a_{1} + (a_{0} + a_{1} x^{2} + . .) + 5 a_{0} = \frac{1}{1 - x}$ $\Rightarrow \frac{1}{x} Λ (x) - \frac{1}{x} - 2 + Λ (x) + 5 = \frac{1}{1 - x}$ $\Rightarrow Λ (x) (\frac{x + 1}{x}) = \frac{1}{x} + \frac{1}{1 - x} - 3$ $\Rightarrow Λ (x) = \frac{1}{1 + x} + \frac{x}{(1 - x) (1 + x)} - \frac{3 x}{(1 + x)} . .$ $W e h a v e t o t u r n i t i n a T a y l o r s e r i e s$
Commented by mr W last updated on 04/May/21

$t h a n k s t o a l l f o r t r y i n g d i f f e r e n t w a y s!$
	Answered by mr W last updated on 04/May/21

	$s u c h t h a t {i t}^{'} s n o t t o o s p e c i a l, i^{'} l l$ $c h a n g e t h e c o n d i t i o n t o$ $a_{0} = 1, a_{1} = 3 .$ $a_{n} - 5 a_{n - 1} + 6 a_{n - 2} = 1$ $a t f i r s t w e s h a l l t r y t o e l i m i n a t e t h e$ $t e r m 1 o n t h e R H S .$ $l e t a_{n} = b_{n} + k, k i s a c o n s t a n t .$ $t h e n w e w i l l g e t$ $b_{n} + k - 5 (b_{n - 1} + k) + 6 (b_{n - 2} + k) = 1$ $b_{n} - 5 b_{n - 1} + 6 b_{n - 2} = 1 - 2 k$ $w e s e t 1 - 2 k = 0, i . e . k = \frac{1}{2}, t h e n$ $b_{n} - 5 b_{n - 1} + 6 b_{n - 2} = 0$ $l e t b_{n} = {A p}^{n}$ ${A p}^{n} - 5 {A p}^{n - 1} + 6 {A p}^{n - 2} = 0$ ${A p}^{n - 2} (p^{2} - 5 p + 6) = 0$ $s i n c e A \neq 0, p \neq 0,$ $\Rightarrow p^{2} - 5 p + 6 = 0$ $\Rightarrow (p - 2) (p - 3) = 0$ $\Rightarrow p = 2 o r 3$ $b_{n} c a n b e g e n e r a l l y e x p r e s s e d a s$ $b_{n} = A 2^{n} + B 3^{n}$ $w i t h A, B = c o n s t a n t s$ $\Rightarrow a_{n} = A 2^{n} + B 3^{n} + \frac{1}{2}$ $a_{0} = A + B + \frac{1}{2} = 1 (g i v e n) . . . (i)$ $a_{1} = A \times 2 + B \times 3 + \frac{1}{2} = 3 (g i v e n) . . . (i i)$ $f r o m (i) a n d (i i) w e g e t$ $\Rightarrow A = - 1$ $\Rightarrow B = \frac{3}{2}$ $\Rightarrow a_{n} = - 2^{n} + \frac{3}{2} \times 3^{n} + \frac{1}{2}$ $\Rightarrow a_{n} = \frac{3^{n + 1} + 1}{2} - 2^{n}$ $◼$
	Commented by BHOOPENDRA last updated on 04/May/21

	$t h a n k s a l o t s i r$
	Answered by floor(10²Eta[1]) last updated on 05/May/21

	$a_{n} = (\frac{5 - \sqrt{5}}{10}) {(\frac{5 + \sqrt{5}}{2})}^{n} + (\frac{5 + \sqrt{5}}{10}) {(\frac{5 - \sqrt{5}}{2})}^{n}$
	Commented by mr W last updated on 05/May/21

	$w r o n g .$ $a_{2} = 15, b u t s h o u l d b e 14 .$
	Commented by floor(10²Eta[1]) last updated on 05/May/21

	$wrong .$ $a_{2} - {5 a}_{1} + {6 a}_{0} = 1$ $a_{2} = 5$
	Commented by mr W last updated on 05/May/21

	$a t y p o . i m e a n t a_{3} s h o u l d b e 14, n o t 15 .$
	Commented by floor(10²Eta[1]) last updated on 05/May/21

	${you}^{'} re right$
	Answered by floor(10²Eta[1]) last updated on 05/May/21

	$a_{n} = x_{n} + y_{n}$ $a_{n} = {5 a}_{n - 1} - {6 a}_{n - 2} + 1$ $x_{n} = {5 x}_{n} - {6 x}_{n} + 1$ $x_{n} = \frac{1}{2}$ $y_{n} = {5 y}_{n - 1} - {6 y}_{n - 2}$ $y^{2} - 5 y + 6 = 0$ $y_{1} = 2, y_{2} = 3$ $y_{n} = c_{1} 2^{n} + c_{2} 3^{n}$ $y_{0} = c_{1} + c_{2} = a_{0} - \frac{1}{2} = \frac{1}{2}$ $y_{1} = {2 c}_{1} + {3 c}_{2} = a_{1} - \frac{1}{2} = \frac{3}{2}$ $c_{2} = \frac{1}{2} - c_{1}$ ${2 c}_{1} + \frac{3}{2} - {3 c}_{1} = \frac{3}{2} \Rightarrow c_{1} = 0 \Rightarrow c_{2} = \frac{1}{2}$ $y_{n} = \frac{3^{n}}{2}$ $\Rightarrow a_{n} = \frac{3^{n} + 1}{2}$
	Commented by mr W last updated on 05/May/21

	$t h a n k s s i r!$
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