Mathc initiation/a239

Sommaire

Installer et compiler ces fichiers dans votre répertoire de travail.

	c00c.c

/* --------------------------------- */
/* save as c00c.c                    */
/* --------------------------------- */
#include     "x_a.h"
#include      "fc.h"
/* --------------------------------- */
int main(void)
{
double n = 4;

 clrscrn();
 
 printf("(n+2)! = (n+2) (n+1)     (n)!     \n" 
        "       = (n+2) (n+1) [n  (n-1)!]         n! = [n (n-1)!]\n\n"
        "       = (n+2) (n+1)  n  (n-1)! \n\n"
        
        "             With   n = %.0f   \n\n" 
        "               (n+2)! = %.0f     \n"
        "  (n+2)(n+1)     (n)! = %.0f     \n"
        "  (n+2)(n+1) n (n–1)! = %.0f \n\n\n",
                    n ,
           F_pls2(n),
         F_pls2_a(n),
         F_pls2_b(n) );
 stop();
 
 return 0;
}
/* --------------------------------- */
/* --------------------------------- */

Exemple de sortie écran :

(n+2)! = (n+2) (n+1)     (n)!     
       = (n+2) (n+1) [n  (n-1)!]         n! = [n (n-1)!]

       = (n+2) (n+1)  n  (n-1)! 

             With   n = 4   

               (n+2)! = 720     
  (n+2)(n+1)     (n)! = 720     
  (n+2)(n+1) n (n–1)! = 720 


 Press return to continue.

Avec n = 4, (n+2)! = (6)! = 720

Ici on part  :
     
       (n+2)! = (n+2)             (n+1)!   
              = (n+2) (n+1)       (n-0)! 
              = (n+2) (n+1) (n-0) (n-1)!  
              = (n+2) (n+1) (n)   (n-1)!    
              = (4+2) (4+1) (4)   (4-1)!
              = (6)   (5)   (4)     (3)!  = 720            
      
Le but de ces manupulations est de simplifier les calculs.
                  
            (n+2)!    (n+2) (n+1) (n) (n-1)! 
    ex :    ------- = ----------------------= (n+2) (n+1) n     
            (n–1)!                    (n–1)!