Đáp án:
$\begin{array}{l}
a)M = 2 + {2^2} + {2^3} + ... + {2^{2017}} + {2^{2018}}\\
\Rightarrow 2M = {2^2} + {2^3} + {2^4} + ... + {2^{2018}} + {2^{2019}}\\
\Rightarrow 2M - M = {2^{2019}} - 2\\
\Rightarrow M = {2^{2019}} - 2\\
b)M = 2 + {2^2} + {2^3} + ... + {2^{2017}} + {2^{2018}}\\
= \left( {2 + {2^2}} \right) + \left( {{2^3} + {2^4}} \right).. + \left( {{2^{2017}} + {2^{2018}}} \right)\\
= 2\left( {1 + 2} \right) + {2^3}\left( {1 + 2} \right) + .. + {2^{2017}}\left( {1 + 2} \right)\\
= 2.3 + {2^3}.3 + .. + {2^{2017}}.3\\
= \left( {2 + {2^3} + ... + {2^{2017}}} \right).3 \vdots 3
\end{array}$
Vậy M chia hết cho 3
