Calculati 3+6+9+12+...2001 va rog sa ma ajutati .


Răspuns :

[tex] \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[/tex]
3+ 6+ 9+12+ ... +2 001=
1+2+3+4+ ... +667=
3·(1+2+3+4+...+667)=
3·667·668:2=
3·667·334=
2 001·334=
668 334