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1+5+5^2=31

5^3+5^4+5^5=5^3.(1+5+5^2)=5^3.31

Dễ r đung ko?

=(1+5+5^2)+...+5^402(1+5+5^2)

=31+...+5^402.31

=31(1+...+5^402) chia hết cho 31 

\(1+5+5^2+...+5^{404}=\left(1+5+5^2\right)+...+\left(5^{400}+5^{401}+5^{402}\right)=31+31.5^3+...+31.5^{400}\)

\(=31\left(1+5^3+5^6+...+5^{400}\right)\)chia hết cho 31

Ta có:\(1+5+5^2+\cdot\cdot\cdot+5^{404}\)

      = \(\left(1+5+5^2\right)+\cdot\cdot\cdot+\left(5^{402}+5^{403}+5^{404}\right)\)

      =   \(\left(1+5+25\right)+\cdot\cdot\cdot+\left(5^{402}\cdot1+5^{402}\cdot5+5^{402}\cdot25\right)\)

      =      \(31+\cdot\cdot\cdot+\left(1+5+25\right)\cdot5^{402}\)

      =       \(31\cdot1+...+31\cdot5^{402}\)

      =        \(31\cdot\left(1+...+5^{402}\right)⋮31\)

 Vậy tổng trên chia hết cho 31

Saito Haijme câu hỏi tương tự nhé

      \(1+5+5^2+...+5^{404}\)

\(=5^3\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{404}\left(1+5+5^2\right)\)

\(=\left(1+5+5^2\right)\left(5^3+5^4+...+5^{403}+5^{404}\right)\)

\(=31.\left(5^3+5^4+...+5^{403}+5^{404}\right)\)

Vậy tổng trên chia hết cho 31

1 + 5 + 5² + .... + 5⁴⁰⁴

= ( 1 + 5 ) + ( 5² + 5³ ) + ... + ( 5⁴⁰³ + 5⁴⁰⁴ )

= 6 + 5² . ( 1 + 5 ) + ... + 5⁴⁰³ . ( 1 + 5 )

=6 + 5² . 6 + ... + 5⁴⁰³ . 6

= 6 . ( 1 + 5² + ... + 5^{403 )}

= 3 . 2 . ( 1 + 5² + .... + 5⁴⁰³ ) chia hét cho 3 

Ta có: 1 + 5 + 5² + .... + 5⁴⁰⁴

= ( 1 + 5 ) + ( 5² + 5³ ) + ... + ( 5⁴⁰³ + 5⁴⁰⁴ )

= 6 + 5² . ( 1 + 5 ) + ... + 5⁴⁰³ . ( 1 + 5 )

=6 + 5² . 6 + ... + 5⁴⁰³ . 6

= 6 . ( 1 + 5² + ... + 5⁴⁰³ ) chia hết cho 3 

Đặt A=1+5+5²+...+5⁴⁰³+5⁴⁰⁴

=(1+5+5²)+...+(5⁴⁰²+5⁴⁰³+5⁴⁰⁴)

=1.(1+5+5²)+...+5⁴⁰².(1+5+5²)

=1.31+...+5⁴⁰².31

=31.(1+...+5⁴⁰²) chia hết cho 31 (đpcm)