Réponse : Bonsoir,
b) On trouve [tex]v_{n}=-\frac{1}{3} \times \left(\frac{1}{5}\right)^{n}[/tex].
Puis:
[tex]v_{n}=\frac{u_{n}-1}{u_{n}+3}\\u_{n}-1=v_{n}(u_{n}+3)\\u_{n}-u_{n}v_{n}=1+3v_{n}\\u_{n}(1-v_{n})=1+3v_{n}\\u_{n}=\frac{1+3v_{n}}{1-v_{n}}[/tex].
Donc:
[tex]u_{n}=\frac{1+3\left(-\frac{1}{3} \times \left(\frac{1}{5}\right)^{n}\right)}{1+\frac{1}{3} \times \left(\frac{1}{5}\right)^{n}}=\frac{1-\left(\frac{1}{5}\right)^{n}}{1+\frac{1}{3} \times \left(\frac{1}{5}\right)^{n}}[/tex].
