Question

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

Answer 1

$3+6+9+12+...+99=3(1+2+3+...+33)=3* \frac{33*34}{2}=3*33*$$17=99*17=1683$

Answer 2

$\displaystyle 3+6+9+12+...+99=3(1+2+3+...+33)=3 \times \frac{33(33+1)}{2} = \\ \\ =3 \times \frac{33 \times 34}{2} =3 \times \frac{1122}{2} =3 \times 561=1683$