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Abipipoune06
@Abipipoune06
May 2019
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Bonjour besoin d aide pour la partie trois de mon dm merci d avance
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scoladan
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Bonjour
1 + (1/2) + (1/2)² + ... + (1/2)ⁿ
= (1/2)⁰ + (1/2)¹ + (1/2)² + .... + (1/2)ⁿ
somme des (n + 1) premiers termes d'une suite géométrique de raison q = 1/2 et de premier terme 1.
Donc :
1 + 1/2 + .... + (1/2)ⁿ = (1 - (1/2)ⁿ⁺¹)/(1 - (1/2))
= (1 - (1/2)ⁿ⁺¹)/(1/2)
= 2 x (1 - (1/2)ⁿ⁺¹)
= 2 - 2x(1/2)ⁿ⁺¹
= 2 - (1/2)ⁿ
⇒ Sn = 4 x (2 - (1/2)ⁿ) = 8 - 4(1/2)ⁿ
2) Quand n → +∞, (1/2)ⁿ → 0
donc lim Sn = 8
Au bout de n heures, quand n → +∞, le mélange contient 8 L d'adjuvant
3) Sn > 7,9
⇔ 8 - 4 x (1/2)ⁿ > 7,9
⇔ 4 x (1/2)ⁿ < 0,1
⇔ 1/2ⁿ < 0,025
⇒ 2ⁿ > 1/0,025
⇔ 2ⁿ > 40
⇒ n x ln(2) > ln(40)
⇔ n > ln(40)/ln(2)
soit n > 5,32
soit à partir de n = 6
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Lista de comentários
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Bonjour1 + (1/2) + (1/2)² + ... + (1/2)ⁿ
= (1/2)⁰ + (1/2)¹ + (1/2)² + .... + (1/2)ⁿ
somme des (n + 1) premiers termes d'une suite géométrique de raison q = 1/2 et de premier terme 1.
Donc :
1 + 1/2 + .... + (1/2)ⁿ = (1 - (1/2)ⁿ⁺¹)/(1 - (1/2))
= (1 - (1/2)ⁿ⁺¹)/(1/2)
= 2 x (1 - (1/2)ⁿ⁺¹)
= 2 - 2x(1/2)ⁿ⁺¹
= 2 - (1/2)ⁿ
⇒ Sn = 4 x (2 - (1/2)ⁿ) = 8 - 4(1/2)ⁿ
2) Quand n → +∞, (1/2)ⁿ → 0
donc lim Sn = 8
Au bout de n heures, quand n → +∞, le mélange contient 8 L d'adjuvant
3) Sn > 7,9
⇔ 8 - 4 x (1/2)ⁿ > 7,9
⇔ 4 x (1/2)ⁿ < 0,1
⇔ 1/2ⁿ < 0,025
⇒ 2ⁿ > 1/0,025
⇔ 2ⁿ > 40
⇒ n x ln(2) > ln(40)
⇔ n > ln(40)/ln(2)
soit n > 5,32
soit à partir de n = 6