Question

d) 3+6+9+.....+2013 e cu suma lui Gauss

Answer 1

$\displaystyle \mathtt{3+6+9+...+2013=3(1+2+3+...+671)=3 \cdot \frac{671(671+1)}{2} =} \\ \\ \mathtt{=3 \cdot \frac{671 \cdot 672}{2} =3 \cdot \frac{450912}{2} =3 \cdot 225456=676368}$