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1D2010
@1D2010
January 2021
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bonjour svp aider moi à faire les deux dernières questions
3/a et 3/b
merci d'avance.
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scoladan
Verified answer
Bonjour,
3)a)
Vn+1² = 1 + 24Un+1
= 1 + 24/16 x [1 + 4Un + √(1 + 24Un)]
Vn² = 1 + 24Un et Vn ≥ 0 ⇒ Vn = √(1 + 24Un) et Un = (Vn² - 1)/24
Donc :
Vn+1² = 1 + 3/2 x [1 + 4(Vn² - 1)/24 + Vn]
= 1 + 3/2 x [1 + Vn²/6 - 1/6 + Vn]
= 1 + 3/2 x [1/6(Vn² + 6Vn + 5)]
= 1 + 1/4(Vn² + 6Vn + 5)
= 1/4(Vn² + 6Vn + 9)
= 1/4(Vn + 3)²
Vn+1 ≥ 0 ⇒ Vn+1 = 1/2(Vn + 3)
Et donc Vn+1 - 3 = 1/2(Vn + 3) - 3
= 1/2 x Vn + 3/2 - 6/2
= 1/2 x Vn - 3/2
= 1/2 x (Vn - 3)
⇒ (Vn - 3) géométrique de raison q = 1/2 et de premier terme V₀ = √(1 + 24U₀) = 5
b) On en déduit Vn - 3 = 5 x (1/2)ⁿ
Donc Vn = 5 x (1/2)ⁿ + 3
Et Un = (Vn² - 1)/24
= [[5 x (1/2)ⁿ + 3]² - 1]/24
= ...je ne crois pas que ce soit utile de développer...
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Lista de comentários
Verified answer
Bonjour,3)a)
Vn+1² = 1 + 24Un+1
= 1 + 24/16 x [1 + 4Un + √(1 + 24Un)]
Vn² = 1 + 24Un et Vn ≥ 0 ⇒ Vn = √(1 + 24Un) et Un = (Vn² - 1)/24
Donc :
Vn+1² = 1 + 3/2 x [1 + 4(Vn² - 1)/24 + Vn]
= 1 + 3/2 x [1 + Vn²/6 - 1/6 + Vn]
= 1 + 3/2 x [1/6(Vn² + 6Vn + 5)]
= 1 + 1/4(Vn² + 6Vn + 5)
= 1/4(Vn² + 6Vn + 9)
= 1/4(Vn + 3)²
Vn+1 ≥ 0 ⇒ Vn+1 = 1/2(Vn + 3)
Et donc Vn+1 - 3 = 1/2(Vn + 3) - 3
= 1/2 x Vn + 3/2 - 6/2
= 1/2 x Vn - 3/2
= 1/2 x (Vn - 3)
⇒ (Vn - 3) géométrique de raison q = 1/2 et de premier terme V₀ = √(1 + 24U₀) = 5
b) On en déduit Vn - 3 = 5 x (1/2)ⁿ
Donc Vn = 5 x (1/2)ⁿ + 3
Et Un = (Vn² - 1)/24
= [[5 x (1/2)ⁿ + 3]² - 1]/24
= ...je ne crois pas que ce soit utile de développer...