Đáp án:
$\begin{array}{l}
A = 2 + {2^2} + {2^3} + ... + {2^{2010}}\\
= \left( {2 + {2^2}} \right) + \left( {{2^3} + {2^4}} \right) + ... + \left( {{2^{2009}} + {2^{2010}}} \right)\\
= 2.\left( {1 + 2} \right) + {2^3}\left( {1 + 2} \right) + ... + {2^{2009}}\left( {1 + 2} \right)\\
= 2.3 + {2^3}.3 + ... + {2^{2009}}.3\\
= \left( {2 + {2^3} + ... + {2^{2009}}} \right).3 \vdots 3\\
Vay\,A \vdots 3\\
A = 2 + {2^2} + {2^3} + ... + {2^{2010}}\\
= \left( {2 + {2^2} + {2^3}} \right) + \left( {{2^4} + {2^5} + {2^6}} \right) + .. + \\
+ \left( {{2^{2008}} + {2^{2009}} + {2^{2010}}} \right)\\
= 2\left( {1 + 2 + {2^2}} \right) + {2^4}\left( {1 + 2 + {2^2}} \right) + ... + \\
+ {2^{2008}}\left( {1 + 2 + {2^2}} \right)\\
= 2.7 + {2^4}.7 + ... + {2^{2008}}.7\\
= \left( {2 + {2^4} + ... + {2^{2008}}} \right).7 \vdots 7\\
Vay\,A \vdots 7
\end{array}$
