\(A=2^1+2^2+2^3+...+2^{60}\\=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^{59}+2^{60})\\=6+2^2\cdot(2+2^2)+2^4\cdot(2+2^2)+...+2^{58}\cdot(2+2^2)\\=6+2^2\cdot6+2^4\cdot6+...+2^{58}\cdot6\\=6\cdot(1+2^2+2^4+...+2^{58})\)
Vì \(6\cdot(1+2^2+2^4+...+2^{58})\vdots6\)
nên \(A\vdots6(dpcm)\)
\(A=2^1+2^2+...+2^{60}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(A=\left(2+4\right)+2^2\cdot\left(2+4\right)+...+2^{58}\cdot\left(2+4\right)\)
\(A=6+2^2\cdot6+...+2^{58}\cdot6\)
\(A=6\cdot\left(1+2^2+...+2^{58}\right)\) ⋮ 6
Vậy A ⋮ 6
`#3107.101107`
`A = 2^1 + 2^2 + 2^3 + ... + 2^60`
`= (2 + 2^2) + (2^3 + 2^4) + ... + (2^59 + 2^60)`
`= (2 + 2^2) + 2^2 * (2 + 2^2) + ... + 2^58 * (2 + 2^2)`
`= (2 + 2^2)*(1 + 2^2 + ... + 2^58)`
`= 6 * (1 + 2^2 + ... + 2^58)`
Vì `6 * (1 + 2^2 + ... + 2^58) \vdots 6`
`=> A \vdots 6`
Vậy, `A \vdots 6.`
