Question

cho A= 1+ 3 + 3 mũ 2 + 3 mũ 3 +......+ 3 mũ 101 chứng minh A chia hết cho 13

Answer 1

 \(A=1+3+3^2+...+3^{101}\)

\(=\left(1+3+3^2\right)+...+\left(3^{99}+3^{100}+3^{101}\right)\)

\(=\left(1+3+3^2\right)+...+3^{99}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{99}\right)⋮13\)

Answer 2

 

Answer 3

A = 1 + 3 + 3² + .... + 3¹⁰¹

= [ 1+3+3² ] + ..... + [ 3⁹⁹ + 3¹⁰⁰ +3¹⁰¹ ]

= [ 1+ 3+ 3² ] + .... + 3⁹⁹  . [ 1+3+3² ]

= 13. [ 1 + 3³ + .... + 3⁹⁹ ] ⋮ 13