Bonjour :))
[tex]\boxed{U_{n+1}=3U_n+2}\\\\U_1=3U_0+2=3*2+2=8\\U_2=26\\U_3=80\\U_4=242\\\\\frac{U_{n+1}-2}{3}=U_n\ \ \Leftrightarrow \ \ \boxed{U_{n-1}=\frac{1}{3}U_n-\frac{2}{3}}[/tex]
[tex](U_n)=\begin{cases}U_0=100\\U_{n+1}=U_n-3\end{cases}\\\\On\ consid\`ere\ (U_n)\ dont\ le\ terme\ initial\ est\ U_0=100\ et\ tel\ que\ tout\\terme\ s'obtient\ en\ retranchant\ 3\ au\ terme\ pr\'ecedent.\\\\U_1=U_0-3=100-3=97\\U_2=94\\U_3=91\\U_4=88\\\\U_{n+1}+3=U_n\ \Leftrightarrow\ \boxed{U_{n-1}=U_n+3}[/tex]
[tex]\boxed{U_{n+1}=3U_n-2}\\\\U_1=3U_0-2=3*41-2=121\\U_2=361\\U_3=1081\\U_4=3241\\\\\frac{U_{n+1}+2}{3}=U_n\ \Leftrightarrow\ \boxed{U_{n-1}=\frac{1}{3}U_n+\frac{2}{3}}[/tex]
[tex]On\ consid\`ere\ la\ suite\ (U_n)\ dont\ le\ terme\ initial\ est\ U_0=0\ et\ tel\ que\\tout\ terme\ s'obtient\ en\ multipliant\ par\ (-4)\ le\ terme\ pr\'ecedent\ puis\\en\ ajoutant\ (+3)[/tex]
[tex]U_1=-4U_1+3=-4*0+3=3\\U_2=9\\U_3=39\\U_4=-153\\\boxed{U_{14}=-4U_{13}+3}\\\\\boxed{U_{n-1}=-\frac{1}{4}U_n+\frac{3}{4}}[/tex]
Bonne continuation :))
