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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 : \( u_{n+1}=u_{n}-3 \).

Ask by Franklin Dunn. in Gabon
Jan 23,2025

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**EXERCICE 3** On considère la suite \( \left(u_{n}\right) \) définie par : \[ u_{n} = 7 - 3n \] ### a) Calcul des quatre premiers termes 1. **Pour \( n = 0 \) :** \[ u_{0} = 7 - 0 = 7 \] 2. **Pour \( n = 1 \) :** \[ u_{1} = 7 - 3 = 4 \] 3. **Pour \( n = 2 \) :** \[ u_{2} = 7 - 6 = 1 \] 4. **Pour \( n = 3 \) :** \[ u_{3} = 7 - 9 = -2 \] Les quatre premiers termes sont : 7, 4, 1, -2. ### b) Démonstration que \( u_{n+1} = u_{n} - 3 \) **Calcul de \( u_{n+1} \) :** \[ u_{n+1} = 7 - 3(n + 1) \] \[ u_{n+1} = 7 - 3n - 3 \] \[ u_{n+1} = u_{n} - 3 \] Ainsi, \( u_{n+1} = u_{n} - 3 \). **Conclusion :** La suite \( \left(u_{n}\right) \) est une suite arithmétique de raison -3.

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Pour calculer les quatre premiers termes de la suite \( \left(u_{n}\right) \), on remplace \( n \) par 0, 1, 2 et 3 dans l'expression \( u_{n}=7-3 n \). - Pour \( n=0 \) : \( u_{0} = 7 - 3 \cdot 0 = 7 \). - Pour \( n=1 \) : \( u_{1} = 7 - 3 \cdot 1 = 4 \). - Pour \( n=2 \) : \( u_{2} = 7 - 3 \cdot 2 = 1 \). - Pour \( n=3 \) : \( u_{3} = 7 - 3 \cdot 3 = -2 \). Les quatre premiers termes de la suite sont donc : \( 7, 4, 1, -2 \). Pour montrer que \( u_{n+1} = u_{n} - 3 \), on calcule \( u_{n+1} \) : \[ u_{n+1} = 7 - 3(n+1) = 7 - 3n - 3 = 4 - 3n. \] On sait que \( u_{n} = 7 - 3n \). En soustrayant 3 de \( u_{n} \) : \[ u_{n} - 3 = (7 - 3n) - 3 = 4 - 3n = u_{n+1}. \] Ainsi, la relation est vérifiée : \( u_{n+1} = u_{n} - 3 \).

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