Question

Chứng tỏ: A=3+3^2+3^3+3^4+3^5+3^6...+3^28+3^29+3^30 chia hết cho 13

Answer 1

A=3+3²+3³...+3²⁹+3³⁰

A=(3+3²+3³)+...+(3²⁸+3²⁹+3³⁰)

A=3.(1+3+3²)+...+3²⁸.(1+3+3²)

A=3.13+...+3²⁸.13

A=13.(3+...+3²⁸) chia hết cho 13

Answer 2

A= 3(1+3+3^2)+3^4(1+3+3^2)+3^7(1+3+3^2)+...+3^28.(1+3+3^2)

A=(1+3+3^2)(3+3^4+3^7+...+3^25+3^28)

=13.(3+3^4+3^7+...3^28) vậy A chia hết cho 13