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3+6+9+12+...+99 multumesc

Răspuns :

[tex]3+6+9+12+...+99=3(1+2+3+...+33)=3* \frac{33*34}{2}=3*33* [/tex][tex]17=99*17=1683[/tex]
[tex]\displaystyle 3+6+9+12+...+99=3(1+2+3+...+33)=3 \times \frac{33(33+1)}{2} = \\ \\ =3 \times \frac{33 \times 34}{2} =3 \times \frac{1122}{2} =3 \times 561=1683[/tex]