math science corrigé
1139 mots
5 pages
CHAPITRE 1Exercices 1.1
8
1.
∑5 = 5 + 5 + 5 + k =1
7
2.
∑ (i
+ 5 = 40
8 termes
2
+ 1) = ( 42 + 1) + ( 52 + 1) + ( 62 + 1) + ( 72 + 1) = 130
i =4
17
3.
+ 35 = ∑ ( 2k + 1)
3+5+7+
k =1
n
4.
∑ (2
i
− 2i −1 ) = ( 21 − 20 ) + ( 22 − 21 ) + ( 23 − 22 ) +
+ ( 2n −1 − 2n − 2 ) + ( 2n − 2n −1 ) = 2n − 20 = 2n − 1
i =1
n
5.
∑ f ( x ) ∆x k k
= f ( x1 ) ∆x1 + f ( x2 ) ∆x2 + f ( x3 ) ∆x3 +
+ f ( xn ) ∆xn
k =1
Exercices 1.2
25
a)
25
k =1
1.
k =1
∑ 4k = 4 ∑ k = 4
25 ( 25 + 1)
= 1300
2
b)
19 k 1 100
1 100
1 100 (100 + 1) 19 (19 + 1)
−
∑ 4 = 4 k∑ k = 4 ∑ k − ∑ k = 4
= 1215
2
2
k = 20
= 20
k =1 k =1
c)
∑ 2k
100
20
k =8
2
20
7
20 ( 20 + 1) [ 2 ( 20 ) + 1] 7 ( 7 + 1) [ 2 ( 7 ) + 1]
20
= 2∑ k 2 = 2 ∑ k 2 − ∑ k 2 = 2
−
= 5 460
6
6
k =1
k =8 k =1
n
2.
∑ ( A + Ck ) = n [ A +
1
2
C ( n + 1)]
k =1
PREUVE n n
n
k =1
k =1
k =1
∑ ( A + Ck ) = ∑ A + C ∑ k n ( n + 1)
2
= n [ A + 12 C ( n + 1)]
= nA + C
© ERPI
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n 2 ( n + 1) k =
∑
4 k =1 n 3.
2
3
PREUVE
On a n
∑ k
4
4
− ( k − 1) = n 4 − 0 4
[Somme télescopique.]
k =1
= n4
De plus, n
∑ k
n
4
4
− ( k − 1) = ∑ [ k 4 − ( k 4 − 4k 3 + 6k 2 − 4k + 1)]
k =1
k =1 n = ∑ ( 4k 3 − 6k 2 + 4k − 1) k =1 n n
n
n
k =1
k =1
k =1
k =1
= 4∑ k 3 − 6∑ k 2 + 4∑ k − ∑ 1 n n
n
n
k =1
k =1
k =1
k =1
de sorte que n 4 = 4∑ k 3 − 6∑ k 2 + 4∑ k − ∑ 1, d’où n ∑k3 = k =1
n n n
1 4
n + 6∑ k 2 − 4∑ k + ∑ 1
4 k =1 k =1 k =1
1 4 n ( n + 1)( 2n + 1) n ( n + 1) n +6
−4
+ n
4
6
2
n
= [ n 3 + 2n 2 + 3n + 1 − 2n − 2 + 1]
4
n
= ( n 3 + 2n 2 + n )
4
n2 ( 2
=
n + 2n + 1)
4
2 n 2 ( n + 1)
=
4
=
© ERPI
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