Question

[Ho Thi Nhu Y]

Cho A = 2 + 2² + 2³ + ...... + 2⁶⁰

Chứng minh rằng A chia hết cho 6

Giúp mình với

Answer 1

[Trần Nguyễn Hoàng An]

\(Tacó:\left(2+2^2\right)\cdot\left(2^3+2^4\right)\cdot...\cdot\left(2^{59}+2^{60}\right)\)

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

A \(⋮\)6 do A \(\div\)6 \(\times\)6=A

Answer 2

[Huyền Nhi]

-  Xét \(A⋮2\)

Ta có :\(A=2+2^2+2^3+....+2^{60}\)

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

Vì \(2⋮2;\left(1+2+2^2+....+2^{59}\right)\inℕ^∗\)

Nên \(2.\left(1+2+2^2+....+2^{59}\right)⋮2\)

Do đó : \(A⋮2\)          \(\left(1\right)\)

- Xét \(A⋮3\)

Ta có : \(A=2+2^2+2^3+.....+2^{60}\)

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

\(=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+.....+2^{59}\left(1+2\right)\)

\(=2.3+2^3.3+2^5.3+.....+2^{59}.3\)

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

Vì \(3⋮3;\left(2+2^3+2^5+....+2^{59}\right)\inℕ^∗\)

Nên \(3.\left(2+2^3+2^5+....+2^{59}\right)⋮3\)            \(\left(2\right)\)

Từ (1) và (2), kết hợp với \(2.3=6;\left(2,3\right)=1\) suy ra  \(A⋮6\)      \(\left(đpcm\right)\)