Question

calculati
a 3+6+9+12+...+2019
b 4+8+12+..+2020
c 5+10+15+20+...+2020
d 7+14+21+...+...+2023

Answer 1

3(1+2+3+4+.....+673)=3x(673+674):2=

4(1+2+2+....+505)=4x(505x506):2=

5(1+2+3+4+....+404)=5x(404x405):2=

7(1+2+3+....+289)=7x(289x290):2=

continuă cu calculele...

Answer 2

a) ...=3*(1+2+3+...+673)=3*(673*674/2) =680403;

b) ...=4*(1+2+3+...+505)=4*(505*506/2) =511060;

c) ...=5*(1+2+3+...+404)=5*(404*405/2) = 409050;

d) ...=7*(1+2+3+...+289)=7*(289*290/2) =293335;