Question

sa se arate ca numerele n=1997+2×(1+2+3+...+1996)si p=1+3+5+...+1997 sunt păstrate perfecte.Generalizare

Answer 1

n= 1997+2 * (1+2+3+...+1996)
n = 1997 + 2 * $\frac{1996*1997}{2}$
n = 1997 + 1996 * 1997
n = 1997 (1 + 1996)
n = 1997 * 1997
=> n = $1997^2$

p = 1+3+5+...+1997

!! Exista sume Gauss pt numere impare, formula fiind:
1 + 3 + 5 + 7 + … + ( 2n – 1 ) = n * n = $n^2$

=> p = 999 * 999 = $999^2$

Answer 2

n=1997+ 2*(1996*1997)/2=1997+1996*1997=1997(1+1996)=1997*1997=1997², p.p, cerinta
generalizare
n+2(1+2+...+n-1) e p.p
n+2*(n-1) (n-1+1)/2= n+ (n-1)n=n + n²-n=n². p.p., cerrinta

1+3+5+...+1997= 1+1+2*1+1+2*2+1+2*3+....1+2*998=
   (1 +1+...+1)999 de 1 +2(1+2+3+...998)= 999 +2*998*999/2=
999+998*999=999(1+998)=999*999=999², p.p.,  cerinta


generalizare
1+3+...+2k+1=(11...+1) adica (k+1) de 1 +2( 1+2+...+k)=
k+1+2* k*(k+1)/2= k+1 + k (k+ 1)= (k+1)(1+k)= (k+1)², p.p. , cerinta