ta có : \(A=2+2^2+2^3+2^4+...+2^{100}\)
\(\Leftrightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+...+\left(2^{99}+2^{100}\right)\)
\(\Leftrightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(\Leftrightarrow A=3\left(2+2^3+2^5+...+2^{99}\right)⋮3\)
\(\Rightarrow\) \(A\) chia hết cho \(3\) (đpcm)
\(A=2+2^2+2^3+2^4+...+2^{100}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(A=2\cdot3+2^3\cdot3+...+2^{99}\cdot3\)
\(A=3\left(2+2^3+...+2^{99}\right)\)
\(\Rightarrow A⋮3\)
Vậy A\(⋮\)3 (đpcm)
