Question

[Trà My Kute]

Cho A = 2+2²+2³+....+2⁶⁰

Chứng minh rằng A chia hết cho 3, 7 và 15

Answer 1

[Nguyễn Thanh Hằng]

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

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

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

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

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

b/Ta có :

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

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

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

\(=2.7+2^3.7+......+2^{58}.7\)

\(=7\left(2+2^3+.....+2^{58}\right)⋮7\left(đpcm\right)\)

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

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+....+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+....+2^{57}\left(1+2+2^2+2^3\right)\)

\(=2.15+2^5.15+......+2^{57}.15\)

\(=15\left(2+2^5+......+2^{57}\right)⋮15\left(đpcm\right)\)