expression of the sequence (a_n ) defined by { ((a_0 >0 , a_1 >0)),((a_(n+2) =((2(−1)^n )/(n+2))−((2(−1)^n (2n+3))/(n+2))a_(n+1) +((n+1)/(n+2))a


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Question Number 206351 by MetaLahor1999 last updated on 12/Apr/24
$expression of the sequence (a_n ) defined by { ((a_0 >0 , a_1 >0)),((a_(n+2) =((2(−1)^n )/(n+2))−((2(−1)^n (2n+3))/(n+2))a_(n+1) +((n+1)/(n+2))a_n )) :}$
$e x p r e s s i o n o f t h e s e q u e n c e (a_{n}) d e f i n e d$ $b y$ ${\begin{cases} a_{0} > 0, a_{1} > 0 \\ a_{n + 2} = \frac{2 {(- 1)}^{n}}{n + 2} - \frac{2 {(- 1)}^{n} (2 n + 3)}{n + 2} a_{n + 1} + \frac{n + 1}{n + 2} a_{n} \end{cases}$
Commented by TheHoneyCat last updated on 12/Apr/24
$⇔ { ((a_0 >0 , a_1 >0)),((a_(n+2) =((2(−1)^n (1−2n−3))/(n+2))a_(n+1) +((n+1)/(n+2))a_n )) :} ⇔ { ((a_0 >0 , a_1 >0)),((a_(n+2) =((4(−1)^(n+1) (n+1))/(n+2))a_(n+1) +((n+1)/(n+2))a_n )) :} ⇔ { ((a_0 >0 , a_1 >0)),((a_(n+2) =((n+1)/(n+2))(4(−1)^(n+1) a_(n+1) +a_n ))) :} It is not a solution in any way... it′s just that I think this reformulation might make it easier... I might continue to work on it tomorow... good luck to anyone attempting.$
$\Leftrightarrow {\begin{cases} a_{0} > 0, a_{1} > 0 \\ a_{n + 2} = \frac{2 {(- 1)}^{n} (1 - 2 n - 3)}{n + 2} a_{n + 1} + \frac{n + 1}{n + 2} a_{n} \end{cases}$ $\Leftrightarrow {\begin{cases} a_{0} > 0, a_{1} > 0 \\ a_{n + 2} = \frac{4 {(- 1)}^{n + 1} (n + 1)}{n + 2} a_{n + 1} + \frac{n + 1}{n + 2} a_{n} \end{cases}$ $\Leftrightarrow {\begin{cases} a_{0} > 0, a_{1} > 0 \\ a_{n + 2} = \frac{n + 1}{n + 2} (4 {(- 1)}^{n + 1} a_{n + 1} + a_{n}) \end{cases}$ $It is not a solution in any way . . .$ ${it}^{'} s just that I think this reformulation might$ $make it easier . . .$ $I might continue to work on it tomorow . . .$ $good luck to anyone attempting .$
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