Correction - UniversitÃ© du Maine

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1). ∑ 1975 ∼ 1984 = 24000(1 + i) 9 + · · · + 24000(1 + i) + 24000 = 24000[1 + (1 + i) + · · · + (1 + i) 9 ] = 24000 × = 373447 1 − (1 + i)10 1 − (1 + i) TD 5 Exercice 1 Exercice 2 Exercice 3 V 1990 = ∑ 1975 ∼ 1984 × (1 + i) 6 = 643745 2). ⇒ 3). V 1990 = K 0 = a (1 + i)n − 1 i n = ln(1+ K 0 i a ) ln(1+i) = 20 K 20 = A 1 + A 2 + · · · + A 20 = A 1 + A 1 (1 + i) + · · · + A i (1 + i) 19 = A 1 1 − (1 + i) 20 1 − (1 + i) = a(1 + i) −20[ 1 − (1 + i)20 ] = 634491 1 − (1 + i) S = V 1990 − K 20 = 9253.49
Excercice 3 1). 1 er méthode : le taux mensuel proportionnel i 1 au taux annuel (a). i 1 = i 12 = 0.01 TD 5 Exercice 1 Exercice 2 Exercice 3
Page 1 and 2: TD 5 Exercice 1 Exercice 2 Outils d
Page 3 and 4: Excercice 1 Rappel : A p, p eme amo
Page 5 and 6: [ (1+i) n −1 ] 2). K 0 = A 1 i av
Page 7 and 8: [ (1+i) n −1 ] 2). K 0 = A 1 i av
Page 9 and 10: Excercice 2 TD 5 Exercice 1 Exercic
Page 11: 1). ∑ 1975 ∼ 1984 = 24000(1 + i
Page 15 and 16: (c). Mois Capital Amortissement Int
Page 21 and 22: TD 5 2). 2 eme méthode : le taux m
Page 23 and 24: (c). TD 5 Mois Capital restant dû
Page 25 and 26: (c). TD 5 Mois Capital Amortissemen
Page 27: TD 5 Exercice 1 Exercice 2 Exercice

Correction - UniversitÃ© du Maine

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