Giải thích các bước giải:
Ta có:
$\begin{array}{l}
A = {2^1} + {2^2} + {2^3} + {2^4} + ... + {2^{59}} + {2^{60}}\\
\Rightarrow 2A = {2^2} + {2^3} + {2^4} + {2^5} + ... + {2^{60}} + {2^{61}}\\
\Rightarrow 2A - A = \left( {{2^2} + {2^3} + {2^4} + {2^5} + ... + {2^{60}} + {2^{61}}} \right) - \left( {{2^1} + {2^2} + {2^3} + {2^4} + ... + {2^{59}} + {2^{60}}} \right)\\
\Rightarrow A = {2^{61}} - {2^1}\\
\Rightarrow A = {2^{61}} - 2
\end{array}$
+) Chứng minh: $A\vdots 3$
Ta có:
$\begin{array}{l}
4 \equiv 1\left( {\bmod 3} \right)\\
\Leftrightarrow {2^2} \equiv 1\left( {\bmod 3} \right)\\
\Rightarrow {\left( {{2^2}} \right)^{30}} \equiv 1\left( {\bmod 3} \right)\\
\Rightarrow {2^{60}} \equiv 1\left( {\bmod 3} \right)\\
\Rightarrow {2^{61}} \equiv 2\left( {\bmod 3} \right)\\
\Rightarrow {2^{61}} - 2 \equiv 0\left( {\bmod 3} \right)\\
\Rightarrow A \equiv 0\left( {\bmod 3} \right)\\
\Rightarrow A \vdots 3
\end{array}$
+) Chứng minh: $A\vdots 7$
$\begin{array}{l}
64 \equiv 1\left( {\bmod 7} \right)\\
\Leftrightarrow {2^6} \equiv 1\left( {\bmod 7} \right)\\
\Rightarrow {\left( {{2^6}} \right)^{10}} \equiv 1\left( {\bmod 7} \right)\\
\Rightarrow {2^{60}} \equiv 1\left( {\bmod 7} \right)\\
\Rightarrow {2^{61}} \equiv 2\left( {\bmod 7} \right)\\
\Rightarrow {2^{61}} - 2 \equiv 0\left( {\bmod 7} \right)\\
\Rightarrow A \equiv 0\left( {\bmod 7} \right)\\
\Rightarrow A \vdots 7
\end{array}$
+) Chứng minh: $A\vdots 13$
$\begin{array}{l}
64 \equiv - 1\left( {\bmod 13} \right)\\
\Leftrightarrow {2^6} \equiv - 1\left( {\bmod 13} \right)\\
\Rightarrow {\left( {{2^6}} \right)^{10}} \equiv {\left( { - 1} \right)^{10}} = 1\left( {\bmod 13} \right)\\
\Rightarrow {2^{60}} \equiv 1\left( {\bmod 13} \right)\\
\Rightarrow {2^{61}} \equiv 2\left( {\bmod 13} \right)\\
\Rightarrow {2^{61}} - 2 \equiv 0\left( {\bmod 13} \right)\\
\Rightarrow A \equiv 0\left( {\bmod 13} \right)\\
\Rightarrow A \vdots 13
\end{array}$
Ta có điều phải chứng minh.
