if-a-n-5a-n-1-6a-n-2-1-and-a-0-1-a-1-2-find-a-n-in-terms-of-n-

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 \dots . 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})

\dots . \underset{a_{n}}{\underset{⏟}{(1 + 3^{0} + 3 + 3^{2} + \dots .)}}

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} + \dots) + 6 (a_{0} + a_{1} x + a_{2} x^{2} + \dots) = \frac{1}{1 - x}

\Rightarrow \frac{1}{x} (a_{0} + a_{1} x + a_{2} x^{2} + \dots) - \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) \dots (i)

a_{1} = A \times 2 + B \times 3 + \frac{1}{2} = 3 (g i v e n) \dots (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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