[tex]1+2+3+...+264+265\times133 = \\ \\ =\dfrac{264\times(264+1)}{2}+ 265\times 133 = \\ \\ =\dfrac{264\times 265}{2}+265\times 133 = \\ \\ = 132\times 265+265\times 133 = \\ \\ = 265\times(132+133) = \\ \\ = 265\times 265 = \\ \\ = 265^\big2 \\ \\ \\ $M-am folosit de suma lui Gauss \rightarrow \boxed{1+2+3+...+n = \dfrac{n\times(n+1)}{2}}[/tex]
