expression-of-the-sequence-a-n-defined-by-a-0-gt-0-a-1-gt-0-a-n-2-2-1-n-n-2-2-1-n-2n-3-n-2-a-n-1-n-1-n-2-a-n-

Question Number 206351 by MetaLahor1999 last updated on 12/Apr/24

e x p r e s s i o n o f t h e s e q u e n c e (a_{n}) d e f i n e d

b y

{\begin{cases} a_{0} > 0, a_{1} > 0 \\ a_{n + 2} = \frac{2 {(- 1)}^{n}}{n + 2} - \frac{2 {(- 1)}^{n} (2 n + 3)}{n + 2} a_{n + 1} + \frac{n + 1}{n + 2} a_{n} \end{cases}

Commented by TheHoneyCat last updated on 12/Apr/24

\Leftrightarrow {\begin{cases} a_{0} > 0, a_{1} > 0 \\ a_{n + 2} = \frac{2 {(- 1)}^{n} (1 - 2 n - 3)}{n + 2} a_{n + 1} + \frac{n + 1}{n + 2} a_{n} \end{cases}

\Leftrightarrow {\begin{cases} a_{0} > 0, a_{1} > 0 \\ a_{n + 2} = \frac{4 {(- 1)}^{n + 1} (n + 1)}{n + 2} a_{n + 1} + \frac{n + 1}{n + 2} a_{n} \end{cases}

\Leftrightarrow {\begin{cases} a_{0} > 0, a_{1} > 0 \\ a_{n + 2} = \frac{n + 1}{n + 2} (4 {(- 1)}^{n + 1} a_{n + 1} + a_{n}) \end{cases}

It is not a solution in any way \dots

{it}^{'} s just that I think this reformulation might

make it easier \dots

I might continue to work on it tomorow \dots

good luck to anyone attempting .