Question

A =3+3²⁺3³+...+3⁵⁹⁺3⁶⁰

Chứng tỏ rằng A chia hết cho 52 

Answer 1

\(A=3+3^2+3^3+....+3^{60}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+....+\left(3^{59}+3^{60}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)

\(=\left(1+3\right)\left(3+3^3+....+3^{59}\right)\)

\(=4\left(3+3^3+....+3^{59}\right)\)\(⋮\)\(4\)

\(A=3+3^2+3^3+...+3^{60}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right)\left(3+3^4+....+3^{58}\right)\)

\(=13\left(3+3^4+...+3^{58}\right)\)\(⋮\)\(13\)

mà  (4;13) = 1

nên  A  chia hết cho 52