[tex]\texttt{Formula -\ \textgreater \ ''n' este 'ultimul numar din sir}\ \boxed{\frac{n(n+1)}{2} } \\ \\ \\(1+2+3+...+56)= \frac{56(56+1)}{2}= \frac{56 \cdot \not57}{\not2} = 28 \cdot57 = \underline{1596} \\ \\ (1+2+3+...+36)= \frac{36(36+1)}{2} = \frac{\not36 \cdot 37}{\not2} = 18 \cdot 37= \underline{666} \\ \\ \texttt{Acum se face scaderea:} 1596-666= \boxed{\boxed{930}}[/tex]
