Question

Chứng tỏ rằng 1+5+5²+...+5⁴⁰²+5⁴⁰³+5⁴⁰⁴ chia hết cho 31

Answer 1

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