Bonjour !
Exercice 130 :
1. a) u0 = 2x0² - 0 - 2 = -2
u1 = 2x1² - 1 - 2 = 2 - 1 - 2 = -1
u2 = 2x2² - 2 - 2 = 8 - 2 - 2 = 4
u3 = 2x3² - 3 - 2 = 18 - 3 - 2 = 13
u4 = 2x4² - 4 - 2 = 32 - 4 - 2 = 26
1. b) u(n+1) = 2x(n+1)² - (n+1) - 2
u(2n) = 2x(2n)² - 2n - 2
u(n) + 1 = 2n² - n - 1
1. c) u(n+1) - u(n) = 2x(n+1)² - (n+1) - 2 - (2n² - n - 2)
= 2x(n²+2n+1) - n - 1 - 2 - 2n² + n + 2
= 2n² + 4n + 2 - n - 1 - 2 - 2n² + n + 2
= 4n + 1 > 0
Donc u(n+1)-u(n) > 0
Donc la suite u(n) est croissante.
1. d) u(10) = 2x10² - 10 - 2 = 188
u(100) = 19 898
u(1000) = 1 998 998
1. e) Il semble que la suite u(n) tende vers l'infini.
2. a) w1 = 2 - 4/w0 = 2 - 4/4 = 2 - 1 = 1
w2 = 2 - 4/w1 = 2 - 4/1 = 2 - 4 = -2
2. b)
w = 4 (on rentre la valeur de w0)
for i in range (0,20) :
w = 2 - 4/w
print(w)
2. c) -2
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Bonjour !
Exercice 130 :
1. a) u0 = 2x0² - 0 - 2 = -2
u1 = 2x1² - 1 - 2 = 2 - 1 - 2 = -1
u2 = 2x2² - 2 - 2 = 8 - 2 - 2 = 4
u3 = 2x3² - 3 - 2 = 18 - 3 - 2 = 13
u4 = 2x4² - 4 - 2 = 32 - 4 - 2 = 26
1. b) u(n+1) = 2x(n+1)² - (n+1) - 2
u(2n) = 2x(2n)² - 2n - 2
u(n) + 1 = 2n² - n - 1
1. c) u(n+1) - u(n) = 2x(n+1)² - (n+1) - 2 - (2n² - n - 2)
= 2x(n²+2n+1) - n - 1 - 2 - 2n² + n + 2
= 2n² + 4n + 2 - n - 1 - 2 - 2n² + n + 2
= 4n + 1 > 0
Donc u(n+1)-u(n) > 0
Donc la suite u(n) est croissante.
1. d) u(10) = 2x10² - 10 - 2 = 188
u(100) = 19 898
u(1000) = 1 998 998
1. e) Il semble que la suite u(n) tende vers l'infini.
2. a) w1 = 2 - 4/w0 = 2 - 4/4 = 2 - 1 = 1
w2 = 2 - 4/w1 = 2 - 4/1 = 2 - 4 = -2
2. b)
w = 4 (on rentre la valeur de w0)
for i in range (0,20) :
w = 2 - 4/w
print(w)
2. c) -2
N'hésite pas si tu as une question :)