Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
A = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{{2^n}}}\\
\Rightarrow 2A = \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + ... + \frac{2}{{{2^n}}}\\
\Leftrightarrow 2A = 1 + \frac{1}{2} + \frac{1}{4} + .... + \frac{1}{{{2^{n - 1}}}}\\
\Rightarrow 2A - A = \left( {1 + \frac{1}{2} + \frac{1}{4} + .... + \frac{1}{{{2^{n - 1}}}}} \right) - \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .... + \frac{1}{{{2^n}}}} \right)\\
\Leftrightarrow A = 1 - \frac{1}{{{2^n}}} = \frac{{{2^n} - 1}}{{{2^n}}}
\end{array}\]
