Bonjour, j'ai un DM de maths a rendre et j'ai un problème sur un question stp: On souhaite monter que 1/n-1/n+1 = 1/n(n+1) où n est un entier naturel non nul 1:

bonjour

1 /n - 1 / ( n + 1 )

⇔ 1 ( n + 1 ) / ( n ( n + 1 )  - 1 ( n ) / n ( n + 1 )

⇔  ( n + 1 - n ) / n ( n + 1 )

⇔   1 / n ( n + 1 )

si n = 1

1/1 ( 1 + 1 ) = 1 / 2

si n = 2

1 / ( 2 ( 2 + 1 ) =  1 / 6

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90

⇔ 9 /10

Bonsoir,

1)

[tex]n=1:\\\\\dfrac{1}{1} -\dfrac{1}{1+1} =1 -\dfrac{1}{2} =\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{1}{1*2}\\\\n=2:\\\\\dfrac{1}{2} -\dfrac{1}{2+1} =\dfrac{1}{2} -\dfrac{1}{3} =\dfrac{3}{6}-\dfrac{2}{3}=\dfrac{1}{6}=\dfrac{1}{2*3}\\\\n=3:\\\\\dfrac{1}{3} -\dfrac{1}{3+1} =\dfrac{1}{3} -\dfrac{1}{4} =\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}=\dfrac{1}{3*4}\\\\[/tex]

2)

[tex]\forall n \in \mathbb{N}_0:\\\\\dfrac{1}{n} -\dfrac{1}{n+1} =\dfrac{n+1}{n(n+1)}-\dfrac{n}{n*(n+1)}=\dfrac{1}{n*(n+1)}\\[/tex]

3)

[tex]\dfrac{1}{2} +\dfrac{1}{6} +\dfrac{1}{12} +\dfrac{1}{20} +\dfrac{1}{30} +\dfrac{1}{42} +\dfrac{1}{56} +\dfrac{1}{72} +\dfrac{1}{90} \\\\=\dfrac{1}{1*2} +\dfrac{1}{2*3} +\dfrac{1}{3*4} +\dfrac{1}{4*5} +\dfrac{1}{5*6} +\dfrac{1}{6*7} +\dfrac{1}{7*8} +\dfrac{1}{8*9} +\dfrac{1}{9*10} \\\\\displaystyle =\sum_{i=1}^9\ \dfrac{1}{i*(i+1)} \\\\=\sum_{i=1}^9\ (\dfrac{1}{i}-\dfrac{1}{i+1}) \\\\[/tex]

[tex]\\=(\dfrac{1}{1}-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+(\dfrac{1}{3}-\dfrac{1}{4})+(\dfrac{1}{4}-\dfrac{1}{5})+(\dfrac{1}{5}-\dfrac{1}{6})+(\dfrac{1}{6}-\dfrac{1}{7})+(\dfrac{1}{7}-\dfrac{1}{8})+(\dfrac{1}{8}-\dfrac{1}{9})+(\dfrac{1}{9}-\dfrac{1}{10})\\\\=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\\\=\dfrac{1}{1}-\dfrac{1}{10}\\\\=\dfrac{10}{10}-\dfrac{1}{10}\\\\=\dfrac{9}{10}\\[/tex]