# math science corrigé

1139 mots 5 pages
CHAPITRE 1
Exercices 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

-1-

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
=