Bonjour Madaadi239,
Hérédité :
Supposons que l'on a : [tex]u_n\ \textgreater \ 2[/tex]
Montrons que [tex]u_{n+1}\ \textgreater \ 2[/tex]
En effet,
[tex]u_{n+1}=\dfrac{4u_n-2}{u_n+1}\\\\u_{n+1}=\dfrac{4u_n+4-6}{u_n+1}\\\\u_{n+1}=\dfrac{4(u_n+1)-6}{u_n+1}\\\\u_{n+1}=\dfrac{4(u_n+1)}{u_n+1}-\dfrac{6}{u_n+1}\\\\\boxed{u_{n+1}=4-\dfrac{6}{u_n+1}}[/tex]
[tex]u_n\ \textgreater \ 2\Longrightarrow u_n+1\ \textgreater \ 2+1\\u_n\ \textgreater \ 2\Longrightarrow u_n+1\ \textgreater \ 3\\\\u_n\ \textgreater \ 2\Longrightarrow \dfrac{1}{u_n+1}\ \textless \ \dfrac{1}{3}\\\\u_n\ \textgreater \ 2\Longrightarrow \dfrac{-6}{u_n+1}\ \textgreater \ \dfrac{-6}{3}\\\\u_n\ \textgreater \ 2\Longrightarrow -\dfrac{6}{u_n+1}\ \textgreater \ -2\\\\u_n\ \textgreater \ 2\Longrightarrow 4-\dfrac{6}{u_n+1}\ \textgreater \ 4-2[/tex]
[tex]\\\\u_n\ \textgreater \ 2\Longrightarrow 4-\dfrac{6}{u_n+1}\ \textgreater \ 2\\\\\boxed{u_n\ \textgreater \ 2\Longrightarrow u_{n+1}\ \textgreater \ 2}[/tex]
L'hérédité est donc vraie.
