Réponse :
Explications étape par étape :
[tex]\left\{\begin{array}{cccc}x_0&=&3&\\y_0&=&1&\\x_{n+1}&=&-2y_n+4&(1)\\y_{n+1}&=&x_n-2&(2)\\\end{array} \right.\\\\\\(2) \Longrightarrow\ y_{n+2}=x_{n+1}-2=-2y_n+4-2=-2y_n+2\\y_{n+2}+2y_n=2\ (3)\\\Longrightarrow\ y_{n+3}+2y_{n+1}=2\ (4)\\(4)-(3)\Longrightarrow\ y_{n+3}-y_{n+2}+2y_{n+1}-2y_{n}=0\\equation\ caract\'eristique\ :\\r^3-r^2+2r-2=0\\r^2(r-1)+2(r-1)=0\\(r-1)(r^2+2)=0\\(r-1)(r-i\sqrt{2})(r+i\sqrt{2})=0\\\\y_n=k_1+k_2(i\sqrt{2})^n+k_3(-i\sqrt{2})^n[/tex]
[tex]x_0\ =\ 3\ ,\ y_0\ =\ 1\\x_1=-2*y_0+4=-2+4=2\\y_1=x_0-2=3-2=1\\\\x_2=-2*y_1+4=2\\y_1=x_1-2=0\\\\x_3=-2*y_2+4=4\\y_3=x_2-2=0\\...\\\\\left\{\begin{array}{cccc}y_0=&k_1+k_2+k_3&=&0\\y_1=&k_1+k_2*i\sqrt{2}+k_3*(-i)\sqrt{2}&=&1\\y_2=&k_1-2*k_2-2*k_3=0\\\end{array} \right.\\\\\\\left\{\begin{array}{ccc}k_1&=&\dfrac{2}{3}\\k_2&=&\dfrac{2-i\sqrt{2}}{12}\\k_3&=&\dfrac{2+i\sqrt{2}}{12}\\\end{array} \right.\\\\\\[/tex]
[tex]s=k_2*(i*\sqrt{2})^n+k_3*((-i)*\sqrt{2})^n=\dfrac{2-i\sqrt{2}*(i\sqrt{2})^n}{12}+\dfrac{2+i\sqrt{2}*(-i\sqrt{2})^n}{12}\\si\ n\ est\ pair:\ n=2p\\s=\dfrac{(-2)^p*(2-i\sqrt{2}+2+i\sqrt{2})}{12}=\dfrac{(-2)^\dfrac{n}{2}}{3}y_n=\dfrac{2+(-2)^\dfrac{n}{2}}{3}si\ n\ est\ impair:\ n=2p+1\\s=\dfrac{(-2)^\dfrac{n-1}{2}}{3}\\\\y_n=\dfrac{2+(-2)^{ \dfrac{n-1}{2} } }{3}\\\\[/tex]
[tex]A\ titre\ d'exercice\, je\ vous\ laisse\ le\ soin\ de\ trouver\ x_n\\x_n=k_1+k_2*(i\sqrt{2})^n+k_3*(-i\sqrt{2})^n\ avec\ k_1=\dfrac{8}{3}\\k_2=\dfrac{1+i\sqrt{2}}{6}\\k_3=\dfrac{1-i\sqrt{2}}{6}\\si\ n\ est\ pair:\ x_n=\dfrac{8+(-2)^{\dfrac{n}{2} } }{3}\\sinon\ \ x_n=\dfrac{8+(-2)^{\dfrac{n+1}{2} } }{3}\\[/tex]
[tex]a)\ voir\ le\ fichier\ xls \ joint\\\\b)\\On\ pose\ v_n=x_{2n}\ en\ perdant \ la \moiti\'e\ de\ la\ suite:\\D\'emonstration\ directe\\Soit\ n=2p\\x_{n+1}=4-2y_n\ \Longrightarrow\ x_{2p+1}=4-2y_{2p}\\y_{n+1}=x_n-2\ \Longrightarrow\ y_{2p+1}=x_{2p}-2\ \Longrightarrow\ y_{2p}=x_{2p-1}-2\\Longrightarrow\ y_{2p-1}=x_{2p-2}-2\\\\\Longrightarrow\ x_{2p}=4-2(x_{2p-2}-2=8-2x_{2p-2}\\Longrightarrow\ x_{2p+2}=8-2x_{2p}\\\\v_n=x_{2n}\\v_{n+1}=x_{2(n+1)}=x_{2n+2}=8-2x_{2n}=8-2*v_n\\v_0=x_0=3\\[/tex]
[tex]On \ pose\ t_n=v_n-\dfrac{8}{3}\\t_{n+1}=v_{n+1}-\dfrac{8}{3}=8-2v_n-\dfrac{8}{3}=\dfrac{16}{3}-2v_n=-2(v_n-\dfrac{8}{3})=-2*t_n\\t_0=v_0-\dfrac{8}{3}=\dfrac{1}{3}\\t_n=\dfrac{1}{3}*(-2)^nv_n=t_n+\dfrac{8}{3}\\v_n=\dfrac{8+(-2)^n}{3}\\D\'emonstration\ par\ r\'ecurrence\\Initialisation:\\Pour\ n=0:\ v_0=x_{2*0}=3\\v_n=\dfrac{8+(-2)^n}{3}\ est\ vrai\\v_{n+1}=x_{2*(n+1)}=8-2*v_n=8-2*\dfrac{8+(-2)^n}{3}=\dfrac{8+(-2)^{n+1}}{3}\ est\ vrai.[/tex]
[tex]c)\\v_n=x_{2n}\Longrightarrow\ v_{n+1}=x_{2n+1}\\w_n=x_{2n+1}=v_{n+1}=\dfrac{8+(-2)^{n+1}}{3}=\dfrac{8+(-2)*(-2)^{n}}{3}\\d)\\M_{2n}=(x_{2n},y_{2n})\\M_{2n+1}=(x_{2n+1},y_{2n+1})\\[/tex]
