1.2 1) a
1
=
12
=
2
a
2
= 1
12
|{z}
a
1 +
1
23
= 1
2 +
1
6
= 2
3
a
3
= 1
12 +
1
23
| {z }
a2
+ 1
34
= 2
3 +
1
12
= 3
4
a
4
= 1
12 +
1
23 +
1
34
| {z }
a
3
+ 1
45
= 3
4 +
1
20
= 4
5
On peut supposer que a
n
= n
n+1 .
2) b
1
= 2
1
=2
b
2
= 34
13
= 4
1
=4=2 2
b
3
=
456
135
= 42
1
=8=2 3
b
4
=
5678
1357
= 28
1
=16=2 4
On s'attend à e que b
n
=2 n
.
3)
1
= 1
2 2
1
= 1
4 1
= 1
3
2
= 1
2 2
1
| {z }
1 +
1
4 2
1
= 1
3 +
1
16 1
= 1
3 +
1
15
= 2
5
3
= 1
2 2
1 +
1
4 2
1
| {z }
2
+ 1
6 2
1
= 2
5 +
1
36 1
= 2
5 +
1
35
= 3
7
4
= 1
2 2
1 +
1
4 2
1 +
1
6 2
1
| {z }
3
+ 1
8 2
1
= 3
7 +
1
64 1
= 3
7 +
1
63
= 4
9
On imagineque
n
= n
2n+1 .
Onremarqueeneetquelesnumérateursformentlasuited'entiersonsé-
utifs1;2;3;4;::: etque les dénominateurs onstituent la suite d'entiers
impairsonséutifs 3;5;7;9;:::
