Réponse :
Bonsoir,
Explications étape par étape
[tex]\left\{\begin{array}{ccc}u_0&=&1\\v_0&=&3\\u_{n+1}&=&\dfrac{u_n+3v_n}{4} \\v_{n+1}&=&\dfrac{3u_n+v_n}{4} \\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}u_0+v_0&=&4\\u_{n+1}+v_{n+1}&=&u_n+v_n \\u_{n+1}-v_{n+1} & = & -\dfrac{1}{2} (u_n-v_n)\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}u_0-v_0&=&-3\\u_{n+1}+v_{n+1}&=&4 \\u_{n+1}-v_{n+1}&=&-3*(\dfrac{-1}{2})^n \\\end{array}\right.\\\\[/tex]
2)
[tex]\left\{\begin{array}{ccc}u_0-v_0&=&-3\\2u_{n+1}&=&4-3 (\dfrac{-1}{2})^n\\2v_{n+1}&=&4+3*(\dfrac{-1}{2})^n \\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}u_0-v_0&=&-3\\u_{n+1}&=&2+3 (\dfrac{-1}{2})^{n+1}\\v_{n+1}&=&2-3*(\dfrac{-1}{2})^{n +1}\\\end{array}\right.\\\\[/tex]
De là, on peut déduire u(n) et v(n)
