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EXERCICE 3 On considère la suite \( \left(u_{n}\right) \) définie par: \( u_{n}=7-3 n \). - Calculer ses quatre premiers termes. - Montrer que : \( u_{n+1}=u_{n}-3 \). Soit la suite \( \left(v_{n}\right) \) telle que : \( v_{0}=2 \) et \( v_{n+1}=3-2 v_{n} \) - Calculer \( v_{1}, v_{2} \) et \( v_{3} \). - Montrer que \( v_{n+2}=4 v_{n}-3 \).

Ask by Hilton Delgado. in Gabon
Jan 23,2025

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### **Exercice 3** **Suite \( \left(u_{n}\right) \):** \[ u_{n} = 7 - 3n \] - **Premiers termes :** - \( u_0 = 7 \) - \( u_1 = 4 \) - \( u_2 = 1 \) - \( u_3 = -2 \) - **Relation de récurrence :** \[ u_{n+1} = u_{n} - 3 \] **Suite \( \left(v_{n}\right) \):** \[ v_{0} = 2 \] \[ v_{n+1} = 3 - 2v_{n} \] - **Termes calculés :** - \( v_1 = -1 \) - \( v_2 = 5 \) - \( v_3 = -7 \) - **Relation de récurrence :** \[ v_{n+2} = 4v_{n} - 3 \]

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Commençons par la suite \( (u_n) \). En remplaçant \( n \) par les valeurs 0, 1, 2 et 3, nous avons : - \( u_0 = 7 - 3 \times 0 = 7 \) - \( u_1 = 7 - 3 \times 1 = 4 \) - \( u_2 = 7 - 3 \times 2 = 1 \) - \( u_3 = 7 - 3 \times 3 = -2 \) Ensuite, pour montrer que \( u_{n+1} = u_n - 3 \), on remplace \( n \) par \( n+1 \): \( u_{n+1} = 7 - 3(n+1) = 7 - 3n - 3 = 4 - 3n = (7 - 3n) - 3 = u_n - 3 \), ce qui est bien ce qu'il fallait démontrer. Passons à la suite \( (v_n) \). Pour calculer les trois premiers termes: - \( v_1 = 3 - 2 \times v_0 = 3 - 2 \times 2 = -1 \) - \( v_2 = 3 - 2 \times v_1 = 3 - 2 \times (-1) = 3 + 2 = 5 \) - \( v_3 = 3 - 2 \times v_2 = 3 - 2 \times 5 = 3 - 10 = -7 \) Pour montrer que \( v_{n+2} = 4v_n - 3 \), on peut exprimer \( v_{n+2} \) à partir de \( v_n \): - On sait que \( v_{n+1} = 3 - 2v_n \), donc \( v_{n+2} = 3 - 2v_{n+1} = 3 - 2(3 - 2v_n) = 3 - 6 + 4v_n = 4v_n - 3 \). Ainsi, nous avons prouvé l'expression désirée.

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