Question

[Nguyễn Vũ Hoàng Long]

A = 2 + 2² + 2³ + ... + 2⁶⁰

a. Xét xem A có ⋮ 3, A có ⋮ 7, A có ⋮ 5 không ?

Answer 1

[HT.Phong (9A5)]

Ta có:

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

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

\(A=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)

\(A=3\cdot\left(2+2^3+2^5+...+2^{59}\right)\)

A chia hết cho 3

____

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

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

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

\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\) 

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\)

Vậy A chia hết cho 5

____

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

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

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

\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(A=7\cdot\left(2+2^4+...+2^{58}\right)\)

Vậy A chia hết cho 7