Giải:
\(1+5+5^2+...+5^{404}\)
\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{402}+5^{403}+5^{404}\right)\)
\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+...+5^{402}\left(1+5+5^2\right)\)
\(=31+5^3.31+...+5^{402}.31\)
\(=31\left(1+5^3+...+5^{402}\right)\)
Mà \(31⋮31\)
Nên \(31\left(1+5^3+...+5^{402}\right)⋮31\)
Vậy \(1+5+5^2+...+5^{404}⋮31\)
Chúc bạn học tốt!
gom (1+5+52)+(53+54+55)+.......+(5402+5403+5404)
=1 (1+5+52)+53 (1+5+52)+.....+5402 (1+5+52)
=1.31 + 53.31 + .....+5402.31
vì các tích đều chia hết cho 31 => 1+5+52+53+54+55+.......+5402+5403+5404\(⋮31\)
1+5+5\(^2\)+...+5\(^{404}\)
(1+5+5\(^2\))+(5\(^3\)+5\(^4\)+5\(^5\))+.....+(5\(^{402}\)+5\(^{403}\)+5\(^{404}\))
=1.(1+5+5\(^2\))+5\(^3\).(1+5+5\(^2\))+.....+5\(^{402}\).(1+5+5\(^2\))
=1+5+5\(^2\).(1+5\(^3\)+.....+5\(^{402}\))
=31.(1+5\(^3\)+.....+5\(^{402}\)) chia hết cho 31
Vậy 1+5+5\(^2\)+...+5\(^{404}\) chia hết cho 31