Đáp án:
$\begin{array}{l}
M = 1 + 4 + {4^2} + {4^3} + ... + {4^{100}}\\
= 1 + \left( {4 + {4^2}} \right) + \left( {{4^3} + {4^4}} \right) + ... + \left( {{4^{99}} + {4^{100}}} \right)\\
= 1 + 4\left( {1 + 4} \right) + {4^3}\left( {1 + 4} \right) + .. + {4^{99}}\left( {1 + 4} \right)\\
= 1 + 4.5 + {4^3}.5 + ... + {4^{99}}.5\\
= 1 + \left( {4 + {4^3} + .. + {4^{99}}} \right).5\\
Do:\left\{ \begin{array}{l}
\left( {4 + {4^3} + .. + {4^{99}}} \right).5 \vdots 5\\
1\,ko\,chia\,hết\,cho\,5
\end{array} \right.\\
\Rightarrow 1 + \left( {4 + {4^3} + .. + {4^{99}}} \right).5\,chia\,5\,dư\,1
\end{array}$
Vậy M ko chia hết cho 5
