Question

[Nguyễn Bảo Minh]

Cho: A = 2 + 2² + 2³ + 2⁴ +....+ 2⁶⁰

Chứng minh rằng:  A⋮3; A⋮5;A⋮7

Thanks!

Answer 1

[HT.Phong (9A5)]

Answer 2

[Võ Ngọc Phương]

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

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

\(A=6+2^2.\left(2+2^2\right)+...+2^{58}.\left(2+2^2\right)\)

\(A=6+2^2.6+...+2^{58}.6\)

\(A=6.\left(1+2^2+...+2^{58}\right)\)

Vì \(6⋮3\) nên \(6.\left(1+2^2+...+2^{58}\right)⋮3\)

Vậy \(A⋮3\)

_________________

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

\(A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(A=30+...+2^{56}.\left(2+2^2+2^3+2^4\right)\)

\(A=30+...+2^{56}.30\)

\(A=30.\left(1+...+2^{56}\right)\)

Vì \(30⋮5\) nên \(30.\left(1+...+2^{56}\right)⋮5\)

Vậy \(A⋮5\)

_________________

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

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

\(A=14+...+2^{57}.\left(2+2^2+2^3\right)\)

\(A=14+...+2^{57}.14\)

\(A=14.\left(1+...+2^{57}\right)\)

Vì \(14⋮7\) nên \(14.\left(1+...+2^{57}\right)⋮7\)

Vậy \(A⋮7\)

\(#WendyDang\)