Réponse :
Bonjour,
Explications étape par étape :
[tex]u_0=2\\u_1=\dfrac{7}{3} \\u_2=\dfrac{26}{9} \\\\u_{n+1}=\dfrac{2}{3} *u_{n}+\dfrac{1}{3}*n +1\\\\\Longrightarrow\ u_{n+2}=\dfrac{2}{3} *u_{n+1}+\dfrac{1}{3}*(n+1) +1\\\\u_{n+2}-\dfrac{5}{3} *u_{n+1}+\dfrac{2}{3}*u_{n}=\dfrac{1}{3} \\\\u_{n+3}-\dfrac{5}{3} *u_{n+2}+\dfrac{2}{3}*u_{n+1}=\dfrac{1}{3} \Longrightarrow\ u_{n+3}-\dfrac{8}{3} *u_{n+2}+\dfrac{7}{3}*u_{n+1}-\dfrac{2}{3}*u_n=0 \\\\\boxed{3*u_{n+3}-8u_{n+2}+7*u_{n+1}-2u_n=0}\\\\r^3-8r^2+7r-2=0 \ ou\ (r-1)^2*(r-\dfrac{2}{3} )=0\\\\[/tex]
[tex]u_n=k_1*1^n+k_2*n*1^n+(\dfrac{2}{3} )^n\\k_1=0,\ k_2=1,\ k_3=2\\\\\boxed{u_n\\=n+2*(\dfrac{2}{3} )^n}[/tex]
