Given-a-0-1-a-n-1-1-2-3a-n-5a-n-2-4-n-0-n-I-find-a-n-

Question Number 184160 by cortano1 last updated on 03/Jan/23

G i v e n {\begin{cases} a_{0} = 1 \\ a_{n + 1} = \frac{1}{2} (3 a_{n} + \sqrt{5 a_{n}^{2} - 4}) \end{cases}

\forall n ⩾ 0, n \in I

f i n d a_{n} .

Commented by mr W last updated on 04/Jan/23

s o l u t i o n s e e Q 184280

Answered by a.lgnaoui last updated on 03/Jan/23

a_{n + 1} = \frac{3 a_{n}}{2} + \frac{1}{2} \sqrt{5 a_{n}^{2} - 4}

\sqrt{5 a_{n}^{2} - 4} = 2 a_{n + 1} - 3 a_{n}

5 a_{n}^{2} - 4 = 4 a_{n + 1}^{2} + 9 a_{n}^{2} - 12 a_{n} a_{n + 1}

4 a_{n + 1}^{2} - 12 a_{n} a_{n + 1} + 4 a_{n}^{2} + 4 = 0

a_{n + 1}^{2} - 3 a_{n} a_{n + 1} + a_{n}^{2} + 1 = 0

{(a_{n + 1} - \frac{3 a_{n}}{2})}^{2} - (\frac{5 a_{n}^{2} - 4}{4})

(a_{n + 1} - \frac{3 a_{n}}{2} - \frac{\sqrt{5 a_{n}^{2} - 4}}{2}) (a_{n + 1} - \frac{3 a_{n}}{2} + \frac{\sqrt{5 a_{n}^{2} - 4}}{2})

{\begin{cases} a_{n + 1} = \frac{1}{2} (3 a_{n} + \sqrt{5 a_{n}^{2} - 4}) \\ a_{n + 1} = \frac{1}{2} (3 a_{n} - \sqrt{5 a_{n}^{2} - 4}) \end{cases}

a_{n + 1} = \frac{3}{2} a_{n} \Rightarrow

a_{n} = \frac{3}{2} a_{n - 1}

Commented by SEKRET last updated on 03/Jan/23

thanks sir . and you

Commented by SEKRET last updated on 03/Jan/23

\sqrt{5 a_{n}^{2} - 4} = 2 a_{n + 1} - 3 a_{n}

5 a_{n}^{2} - 4 = 4 a_{n + 1}^{2} + 9 a_{n}^{2} - 12 a_{n} a_{n + 1}

4 a_{n + 1}^{2} (+) 4 a_{n}^{2} - 12 a_{n} a_{n + 1} + 4 = 0

Commented by a.lgnaoui last updated on 03/Jan/23

y e s l o o k a t r e c t i f i c a t i o n

t h a n k s; h a p p y n e w y e a r

Commented by SEKRET last updated on 03/Jan/23

your answer is wrong

Answered by SEKRET last updated on 03/Jan/23

a_{n} = a_{n - 1} + \underset{k = 0}{\sum^{n - 1}} a_{k}

Answered by SEKRET last updated on 03/Jan/23

3 a_{n} = a_{n - 1} + a_{n + 1}

Answered by aleks041103 last updated on 03/Jan/23

a_{n + 1} = \frac{- (- 3 a_{n}) + \sqrt{{(- 3 a_{n})}^{2} - 4 a_{n}^{2} - 4}}{2}

a_{n + 1} = \frac{- (- 3 a_{n}) + \sqrt{{(- 3 a_{n})}^{2} - 4.1 . (a_{n}^{2} + 1)}}{2.1}

\Rightarrow a_{n + 1} i s a s o l u t i o n t o

x^{2} - 3 a_{n} x + a_{n}^{2} + 1 = 0

\Rightarrow a_{n + 1}^{2} + a_{n}^{2} - 3 a_{n + 1} a_{n} + 1 = 0

i^{'} l l c o n t i n u e l a t e r