Đáp án: $A = \dfrac{{{2^{2013}} - 1}}{{{2^{2012}}}}$
Giải thích các bước giải:
$\begin{array}{l}
A = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + ... + \dfrac{1}{{{2^{2012}}}}\\
\Rightarrow 2.A = 2 + 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + ... + \dfrac{1}{{{2^{2011}}}}\\
\Rightarrow 2A - A = 2 - \dfrac{1}{{{2^{2012}}}}\\
\Rightarrow A = 2 - \dfrac{1}{{{2^{2012}}}} = \dfrac{{{2^{2013}} - 1}}{{{2^{2012}}}}
\end{array}$
