Răspuns :
3+6+9+.....+90=3(1+2+3+............+30)=3×[(31×30):2]=1395
(1+2+3+............+30)= suma Gauss cu formala de calcul S=[n(n+1)]:2
Se observa ca sirul creste din 3 in 3, si numerele se impart la 3:
[tex]S=3\cdot1+3\cdot2+3\cdot3+...+3\cdot30\\ S=3(1+2+3+..+30)=3\cdot\frac{30\cdot31}{2}=1395[/tex]
[tex]S=3\cdot1+3\cdot2+3\cdot3+...+3\cdot30\\ S=3(1+2+3+..+30)=3\cdot\frac{30\cdot31}{2}=1395[/tex]