Question

3+6+9+12+.....+2001
care ma ajuta ?

Answer 1

$\displaystyle 3+6+9+12+...+2001=3(1+2+3+...+667)= \\ \\ =3 \times \frac{667(667+1)}{2} =3 \times \frac{667 \times 668}{2} = \\ \\ =3 \times \frac{445556}{2} =3 \times 222778=668334$