Bonsoir,
Explications étape par étape
1) voir fichier joint
2)
[tex]\left\{\begin{array}{ccc}u_0&=&1\\\\v_0&=&2\\v_{n+1}&=&\dfrac{2u_n+3v_n}{5} \\\end{array}\right.\\\\2a)\\d_n=v_n-u_n\\\\d_{n_+1}=v_{n+1}-u_{n+1}\\=\dfrac{2u_n+3v_n}{5} -\dfrac{3u_n+2v_n}{5} \\=\dfrac{-u_n+v_n}{5}\\\\\\\boxed{d_{n+1}=\dfrac{1}{5}d_n}\\\\d_0=v_0-u_0=2-1=1\\raison=\dfrac{-u_n+v_n}{5}\\\\2b)\\\boxed{d_{n}=(\dfrac{1}{5})^n}\\\\[/tex]
3a) voir fichier excel.
[tex]s_n=u_n+v_n\\\\s_{n+1}=u_{n+1}+v_{n+1}\\=\dfrac{3u_n+2v_n}{5} +\dfrac{2u_n+3v_n}{5} \\=u_n+v_n\\=s_n\\\\\boxed{s_{n+1}=s_n}\\\\s_0=1+2=3\\\\(s_n)\ est\ constante.\\\\4)\\\left\{\begin{array}{ccc}s_n&=&3\\d_n&=&(\dfrac{1}{5})^n\\ \end{array}\right.\\\\\\\left\{\begin{array}{ccc}u_n+v_n&=&3\\-u_n+v_n&=&(\dfrac{1}{5})^n\\ \end{array}\right.\\\\[/tex]
[tex]\left\{\begin{array}{ccc}u_n&=&\dfrac{3-(\dfrac{1}{5})^n}{2} \\v_n&=&\dfrac{3+(\dfrac{1}{5})^n}{2} \\\end{array}\right.\\[/tex]
5a)
[tex]\left\{\begin{array}{ccc}T_n&=&\sum_{i=0}^n u_i\\\\W_n&=&\sum_{i=0}^n v_i\\\\\end{array}\right.[/tex]
[tex]\left\{\begin{array}{ccc}T_n&=&\sum_{i=0}^n \dfrac{3-(\dfrac{1}{5})^i}{2}\\W_n&=&\sum_{i=0}^n \dfrac{3+(\dfrac{1}{5})^i}{2}\\\end{array}\right.\\\\[/tex]
[tex]\left\{\begin{array}{ccc}T_n&=&\dfrac{3}{2}*(n+1) +\dfrac{5}{8}*((\dfrac{1}{5})^{n+1} -1)\\\\W_n&=&\dfrac{3}{2}*(n+1) -\dfrac{5}{8}*((\dfrac{1}{5})^{n+1} -1)\end{array}\right.\\\\[/tex]
