Bonsoir,
[tex]u_0=0\\
u_{n+1}= -\dfrac{u_n}{3} +4\\
u_0=0\\
u_1=0+4=4=3+1=3+( \frac{-1}{3}) ^0\\
u_2=- \dfrac{4}{3}+4= \dfrac{8}{3}=3+( \frac{-1}{3}) ^1\\
u_3=- \dfrac{8}{3^2}+4= \dfrac{28}{9}=3+( \frac{-1}{3}) ^2\\
u_4=- \dfrac{28}{3^3}+4= \dfrac{80}{27}=3+( \frac{-1}{3}) ^3\\
...\\
\boxed{u_{n}=3+( \frac{-1}{3}) ^{n-1}}\\
[/tex]
2)
Conjecture :
[tex]u_{n}=3+( \frac{-1}{3}) ^{n-1}\\
Initialisation \\
u_1=4=3+( \frac{-1}{3}) ^{0} \ est\ v\'erifi\'ee.\\ \\
H\'er\'edit\'e:\\
On \ suppose\ que\ u_{n}=3+( \frac{-1}{3}) ^{n-1} \est \ vraie\\
u_{n+1}=- \frac{1}{3} u_n+4 =- \frac{1}{3}*(3+( \frac{-1}{3}) ^{n-1} )+4\\
=-1+(\frac{-1}{3}) ^{n}+4\\\\
\boxed{u_{n+1}=3+(\frac{-1}{3}) ^{n}}\\
[/tex]
