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Suites numériques
Expression explicite d'une suite et calculs de ses premiers termes - Exercice 3
10 min
20
C
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P
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C
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‾
:
C
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{\color{red}\underline{COMPETENCES}\;:\;Calculer.}
COMPETENCES
:
C
a
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c
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.
Question 1
Soit
n
n
n
un entier naturel.
Calculer les trois premiers termes de chacune des suites suivantes :
u
n
=
1
3
n
2
−
6
n
+
1
u_{n} =\frac{1}{3}n^{2} -6n+1
u
n
=
3
1
n
2
−
6
n
+
1
Correction
u
0
=
1
3
×
0
2
−
6
×
0
+
1
u_{0} =\frac{1}{3}\times 0^{2} -6\times 0+1
u
0
=
3
1
×
0
2
−
6
×
0
+
1
donc
u
0
=
1
u_{0} =1
u
0
=
1
u
1
=
1
3
×
1
2
−
6
×
1
+
1
u_{1} =\frac{1}{3}\times 1^{2} -6\times 1+1
u
1
=
3
1
×
1
2
−
6
×
1
+
1
u
1
=
1
3
−
6
+
1
u_{1} =\frac{1}{3} -6+1
u
1
=
3
1
−
6
+
1
u
1
=
1
3
−
6
×
3
1
×
3
+
1
×
3
1
×
3
u_{1} =\frac{1}{3} -\frac{6\times 3}{1\times 3}+\frac{1\times 3}{1\times 3}
u
1
=
3
1
−
1
×
3
6
×
3
+
1
×
3
1
×
3
u
1
=
1
−
18
+
3
3
u_{1} =\frac{1-18+3}{3}
u
1
=
3
1
−
18
+
3
donc
u
1
=
−
14
3
u_{1} =-\frac{14}{3}
u
1
=
−
3
14
u
2
=
1
3
×
2
2
−
6
×
2
+
1
u_{2} =\frac{1}{3}\times 2^{2} -6\times 2+1
u
2
=
3
1
×
2
2
−
6
×
2
+
1
u
2
=
4
3
−
12
+
1
u_{2} =\frac{4}{3} -12+1
u
2
=
3
4
−
12
+
1
u
2
=
4
3
−
12
×
3
1
×
3
+
1
×
3
1
×
3
u_{2} =\frac{4}{3} -\frac{12\times 3}{1\times 3}+\frac{1\times 3}{1\times 3}
u
2
=
3
4
−
1
×
3
12
×
3
+
1
×
3
1
×
3
u
2
=
4
−
36
+
3
3
u_{2} =\frac{4-36+3}{3}
u
2
=
3
4
−
36
+
3
donc
u
2
=
−
29
3
u_{2} =-\frac{29}{3}
u
2
=
−
3
29
Question 2
u
n
=
n
+
6
4
n
−
8
u_{n} =\frac{n+6}{4n-8}
u
n
=
4
n
−
8
n
+
6
Correction
u
0
=
0
+
6
4
×
0
−
8
u_{0} =\frac{0+6}{4\times 0-8}
u
0
=
4
×
0
−
8
0
+
6
donc
u
0
=
−
6
8
=
−
3
4
u_{0} =-\frac{6}{8}=-\frac{3}{4}
u
0
=
−
8
6
=
−
4
3
u
1
=
1
+
6
4
×
1
−
8
u_{1} =\frac{1+6}{4\times 1-8}
u
1
=
4
×
1
−
8
1
+
6
donc
u
1
=
−
7
4
u_{1} =-\frac{7}{4}
u
1
=
−
4
7
u
2
=
2
+
6
4
×
2
−
8
u_{2} =\frac{2+6}{4\times 2-8}
u
2
=
4
×
2
−
8
2
+
6
donc
u
2
=
−
8
0
=
0
u_{2} =-\frac{8}{0}=0
u
2
=
−
0
8
=
0
Question 3
u
n
=
5
n
−
2
4
n
u_{n} =\frac{5n-2}{4^{n}}
u
n
=
4
n
5
n
−
2
Correction
u
0
=
5
×
0
−
2
4
0
u_{0} =\frac{5\times 0-2}{4^{0} }
u
0
=
4
0
5
×
0
−
2
donc
u
0
=
−
2
u_{0} =-2
u
0
=
−
2
. On rappelle que
4
0
=
1
4^{0}=1
4
0
=
1
u
1
=
5
×
1
−
2
4
1
u_{1} =\frac{5\times 1-2}{4^{1} }
u
1
=
4
1
5
×
1
−
2
donc
u
1
=
3
4
u_{1} =\frac{3}{4}
u
1
=
4
3
u
2
=
5
×
2
−
2
4
2
u_{2} =\frac{5\times 2-2}{4^{2} }
u
2
=
4
2
5
×
2
−
2
donc
u
2
=
8
16
=
1
2
u_{2} =\frac{8}{16}=\frac{1}{2}
u
2
=
16
8
=
2
1
Question 4
u
n
=
1
4
n
+
7
u_{n} =\frac{1}{4}\sqrt{n+7}
u
n
=
4
1
n
+
7
Correction
u
0
=
1
4
2
×
0
+
7
u_{0} =\frac{1}{4}\sqrt{2\times 0+7}
u
0
=
4
1
2
×
0
+
7
donc
u
0
=
7
4
u_{0} =\frac{\sqrt{7}}{4}
u
0
=
4
7
u
1
=
1
4
2
×
1
+
7
u_{1} =\frac{1}{4}\sqrt{2\times 1+7}
u
1
=
4
1
2
×
1
+
7
donc
u
1
=
9
4
=
3
4
u_{1} =\frac{\sqrt{9}}{4} =\frac{3}{4}
u
1
=
4
9
=
4
3
u
2
=
1
4
2
×
2
+
7
u_{2} =\frac{1}{4}\sqrt{2\times 2+7}
u
2
=
4
1
2
×
2
+
7
donc
u
2
=
11
4
u_{2} =\frac{\sqrt{11}}{4}
u
2
=
4
11