Ta có:
$\begin{array}{l}
A = 2 + {2^2} + {2^3} + {2^4} + ... + {2^{2016}}\\
A = \left( {2 + {2^2}} \right) + \left( {{2^3} + {2^4}} \right) + ... + \left( {{2^{2015}} + {2^{2016}}} \right)\\
A = 2\left( {1 + 2} \right) + {2^3}\left( {1 + 2} \right) + ... + {2^{2015}}\left( {1 + 2} \right)\\
A = 2.3 + {2^3}.3 + ... + {2^{2015}}.3\\
A = \left( {2 + {2^3} + ... + {2^{2015}}} \right).3
\end{array}$
$ \Rightarrow A \vdots 3$ (1)
Mặt khác, ta cũng có:
${A = \left( {2 + {2^2} + {2^3}} \right) + \left( {{2^4} + {2^5} + {2^6}} \right) + ... + \left( {{2^{2014}} + {2^{2015}} + {2^{2016}}} \right)}$
${A = 2\left( {1 + 2 + 4} \right) + {2^4}\left( {1 + 2 + 4} \right) + ... + {2^{2014}}\left( {1 + 2 + 4} \right)}$
${A = 2.7 + {2^4}.7 + ... + {2^{2014}}.7}$
$A = \left( {2 + {2^4} + ... + {2^{2014}}} \right).7$
$ \Rightarrow A \vdots 7$ (2)
Từ (1) và (2) ta suy ra $A \vdots 21$ (đpcm)
