Question

Tổng sau có chia hết cho 3 không?Vì sao?

A=2+2²+2³+........+2²⁰¹⁷+2²⁰¹⁸

Answer 1

\(A=2+2^2+2^3+2^4+...+2^{2017}+2^{2018}.\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2017}+2^{2018}\right).\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2017}\left(1+2\right).\)

\(A=2.3+2^3.3+...+2^{2017}.3.\)

\(A=\left(2+2^3+...+2^{2017}\right).3⋮3\left(đpcm\right).\)

Answer 2

Tổng đó có chia hết cho 3 vì:

A=2+\(2^2+2^3+...+2^{2017}+2^{2018}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2017}+2^{2018}\right)\)

=2.(1+2)+\(2^3.\left(1+2\right)\)+...+\(2^{2017}.\left(1+2\right)\)

=2.3+2\(^3\).3+...+2\(^{2017}\).3

=3.(2+2\(^3\)+...+2\(^{2017}\))\(⋮3\)

Vậy A \(⋮\) 3