Question

Determinați număr natural n astfel încât 1+3+5...(2n-1)+(2n+1)=625

Answer 1

[1+2+3+4+5+...+(2n-1)+2n+(2n+1)]-(2+4+6+...+2n)=625
[1+2+3+...+(2n+1)]-2(1+2+3+...+n)=625
(2n+1)×(2n+2):2 - 2×n×(n+1):2=625
(2n+1)×(2n+2):2 - n (n+1)=625
(2n+1)×2 (n+1):2 - n (n+1)=625
(2n+1)×(n+1) - n (n+1)=625
(n+1)×(2n+1-n)=625
(n+1)×(n+1)=625
(n+1)la puterea 2=625
n+1= √625
n+1=25
n=24