Bonsoir ;
Exercice n° 3.15 .
a)
[tex]u_1 = \dfrac{1}{4} ; u_2 = \dfrac{1}{28} ; u_3 = \dfrac{1}{70} ; u_4 = \dfrac{1}{130} . [/tex]
b)
On a :
[tex] \dfrac{1}{3n-2} - \dfrac{1}{3n+1} = \dfrac{3n+1-3n+2}{(3n-2)(3n+1)} [/tex]
[tex]= \dfrac{3}{(3n-2)(3n+1}} = 3 u_n ,[/tex]
donc :
[tex]3 u_1 = 1 - \not{\dfrac{1}{4}}[/tex]
[tex]3 u_2 = \not{\dfrac{1}{4}} - \not{\dfrac{1}{7}}[/tex]
[tex]3 u_3 = \not{\dfrac{1}{7}} - \not{\dfrac{1}{10}}[/tex]
[tex]3 u_4 = \not{\dfrac{1}{10}} - \not{\dfrac{1}{13}}[/tex]
....
....
[tex]3 u_n = \not{\dfrac{1}{4}} - \dfrac{1}{3n+1}[/tex]
donc :
[tex]3S_n = 1 - \dfrac{1}{3n+1} = \dfrac{3n}{3n+1} \Rightarrow 3S_n = \dfrac{n}{3n+1} .[/tex]
