Salut,
Fiecare termen al sumei poate fi scris așa (k ia valori de la 1 la 50, deci valori pozitive):
[tex]
\left[\sqrt{k\cdot(k+1)}\right]=\left[\sqrt{k^2+k}\right].\ Avem\ c\breve{a}:\\\\k^2<k^2+k<k^2+2k+1\Rightarrow k^2<k^2+k<(k+1)^2\Rightarrow k<\sqrt{k^2+k}<k+1.\\\\De\ aici\ rezult\breve{a}\ c\breve{a}\left[\sqrt{k^2+k}\right]=k,\ unde\ k\in\{1,2,\ldots\,50\}.\\\\Deci\ suma\ este\ S=1+2+\ldots+50=\dfrac{50\cdot51}2=1275.[/tex]
Green eyes.
