recurrence.py

Created by anaellebaecher

Created on October 17, 2023

562 Bytes

Soit nEN et Pn:"2^n>n"
1)INITIALISTAION

. on verifie lexpression pour le
1er term cad n=1
-> 2^n>n  si n=1
   2^1>1  donc P1 est vraie
   
2)HEREDITE

. supposons quil existe un entier
  k tel que Pk vraie:"2^k>k"
. demontrons que Pk+1 est vraie:
  "2^(k+1)>k+1"
SI: 2^k>k      ]x2
ALORS: 2x2^k>2xk
       2^(k+1)>2k  or 2k=k+k>k+1
       2^(k+1)>k+1      car k>1
Pk+1 est vraie 
et 
Pn est hereditaire

3)CONCLU
. P0 est vraie
. Pn est hereditaire
->donc dapres le principe de
raisonnement par recurrence
Pn est vraie pour tout nEN:
  "2^n>n"

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