A = 2 + 22 + 23 + ... + 210
= (2 + 22) + (23 + 24) + ... + (29 + 210)
= 2(1 + 2) + 23(1 + 2) + ... + 29(1 + 2)
= 3(2 + 23 + ... + 29) \(⋮\) 3
Vậy, A \(⋮\) 3
\(A=2+2^2+2^3+.....+2^{10}\)
\(\Leftrightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+.....+\left(2^9+2^{10}\right)\)
\(\Leftrightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+.....+2^9\left(1+2\right)\)
\(\Leftrightarrow A=2.3+2^3.3+.....+2^9.3\)
\(\Leftrightarrow A=3\left(2+2^3+.....+2^9\right)⋮3\)
Vậy \(A⋮3\)
Ta có: 2 + 22 + 23 + ... + 210
= (2 + 22) + (23 + 24) + ... + (29 + 210)
= 2(1+2) + 23(1+2) + ... + 29(1+2)
= 2.3 + 23.3 + ... + 29.3
= 3(2 + 23 + ... + 29)
Vì 3(2 + 23 + ... + 29) ⋮ 3
Nên (2 + 22 + 23 + ... + 210) ⋮ 3
Vậy A ⋮ 3