Question

rezolvati $1^{2} +2^{2} +3^{2} +4^{2}+...n^{2}\ \textless \ \frac{ (2n+1)^{2} }{6}$

Answer 1

$\frac{n(n+1)(2n+1)}{6}$ < $\frac{ (2n+1)^{2} }{6}$

n(n+1)(2n+1) < $(2n+1)^{2}$

$-2 n^{3}+ n^{2} + 3n + 1 \ \textgreater \ 0$

Care este adevarat pentru oricare n<0

Solutia: n ∈ (-∞ , 0)