[tex]\displaystyle 3+6+9+.......+90= \\ \\ \texttt{Factor comun pe 3} \\ \\ 3(1+2+3+...+30)= \\ \\ \texttt{Facem Gaus in paranteza} \\ \\ \frac{n(n+1)}{2}= \frac{30(30+1)}{2}= \frac{30\cdot31}{2}=465 \\\\ \texttt{n=ultimul nr din sir} \\ \\ 3\cdot465=1395[/tex]
