Question

3+6+9+12+......+2001=
5+10+15+.....+2000=
4+8+12+.....+2000=
7+14+21+.....+2002=

Answer 1

$\displaystyle a).3+6+9+12+...+2001=3(1+2+3+...+667)= \\ \\ =3 \times \frac{667(667+1)}{2} =3 \times \frac{667 \times 668}{2} =3 \times \frac{445556}{2} = \\ \\ =3 \times 222778=668334 \\ \\ b).5+10+15+...+2000=5(1+2+3+...+400)= \\ \\ =5 \times \frac{400(400+1)}{2} =5 \times \frac{400 \times 401}{2} =5 \times \frac{160400}{2} = \\ \\ =5 \times 80200=401000$

$\displaystyle c).4+8+12+...+2000=4(1+2+3+...+500)= \\ \\ =4 \times \frac{500(500+1)}{2} =4 \times \frac{500 \times 501}{2} =4 \times \frac{250500}{2} = \\ \\ =4 \times 125250=501000 \\ \\ d).7+14+21+...+2002=7(1+2+3+...+286)= \\ \\ =7 \times \frac{286(286+1)}{2} =7 \times \frac{286 \times 287}{2} =7 \times \frac{82082}{2} = \\ \\ =7 \times 41041=287287$