ĐẶt A=3+3^2+3^3+....+3^100
A= 3(1+3+3^2+3^3) +3^5(1+3+3^2+3^3)+...... + 3^97 ( 1 + 3 + 3^2 + 3^3)
A=3.40 +3^5.40+.....+3^97.40
Vì 40 chia hết cho 40 => 3.40 +3^5.40+.....+3^97.40
Vậy A chia hết cho 40.
A=3+32+33+34+...+3100 chia hết cho 40
A=(3+32+33+34)+(35+36+37+38)+...+(397+398+399+3100)
A=3.(1+3+32+33)+35.(1+3+32+33)+...+397.(1+3+32+33)
A=3.40+35.40+...+397.40
A=40.(3+35+...+397) chia hết cho 40 (đpcm)