a)A=3+3^2+3^3+3^4+...+3^2007 chia hết cho 13

=3\ti$mes$(1+3+3^2)+...+3^2005$\times$(1+3+3^2)

=3$\times$13+...+3^2005$\times$13
=13$\times$(3+...+3^2005)$\vdots$13
vậy A$\vdots$13
b) B=5+5^2+5^3+5^4+...+5^2010 chia hết cho 31
B=5\ti$mes$(1+5+5^2)+...+5^2008$\times$(1+5+5^2)
B=5$\times$31+...+5^2008$\times$31
B=31$\times$(5+...+5^2008)$\vdots$31
Vậy B$\vdots$31
c)C=8+8^2+8^3+8^4+...+8^2020 chia hết cho 9
C=8$\times$(1+8)+...+8^2019$\times$(1+8)
C=8$\times$9+...+8^2019$\times$9
C=9$\times$(8+...+8^2019)$\vdots$9
Vậy C$\vdots$9
d)D=6+6^2+6^3+6^4+...+6^2012 chia hết cho 256
=6$\times$(1+6+6^2+6^3)+...+6^2009$\times$(1+6+6^2+6^3)
=6$\times$256+...+6^2009$\times$256
=256$\times$(6+6^2009)$\vdots$256
Vậy D$\vdots$256
e)E=7+7^2+7^3+7^4+...+7^2024 chia hết cho 400
=7$\times$(1+7+7^2+7^3)+....+7^397$\times$(1+7+7^2+7^3)
=7$\times$400+...7^397$\times$400
=400$\times$(7+...+7^397)$\vdots$400
Vậy E$\vdots$400

a) A = 3·(1+3+$3^{2}$)+$3^{4}$·(1+3+$3^{2}$)+...+$3^{2005}$·(1+3+$3^{2}$)

=(1+3+$3^{2}$)·(3+$3^{4}$+...+$3^{2005}$)

=13·(3+$3^{4}$+...+$3^{2005}$) chia hết cho 13 (đpcm)

b) B = 5·(1+5+$5^{2}$)+$5^{4}$·(1+5+$5^{2}$)+...+$5^{2008}$·(1+5+$5^{2}$)

=(1+5+$5^{2}$)·(5+$5^{4}$+...+$5^{2008}$)

= 31·(5+$5^{4}$+...+$5^{2008}$) chia hết cho 31 (đpcm)

Câu c), d), e) bạn làm tương tự 

Gợi ý: 1+8=9

1+6+6²+6³=259

1+7+7²+7³=400