Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + ... + \frac{{2014}}{{{4^{2014}}}}\\
\Rightarrow 4A = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + .... + \frac{{2014}}{{{4^{2013}}}}\\
\Rightarrow 4A - A = \left( {1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + .... + \frac{{2014}}{{{4^{2013}}}}} \right) - \left( {\frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + ... + \frac{{2014}}{{{4^{2014}}}}} \right)\\
\Leftrightarrow 3A = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + .... + \frac{1}{{{4^{2013}}}} - \frac{{2014}}{{{4^{2014}}}}\\
B = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + .... + \frac{1}{{{4^{2013}}}}\\
\Rightarrow 4B = 4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + .... + \frac{1}{{{4^{2012}}}}\\
\Rightarrow 4B - B = \left( {4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + .... + \frac{1}{{{4^{2012}}}}} \right) - \left( {1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + .... + \frac{1}{{{4^{2013}}}}} \right)\\
\Leftrightarrow 3B = 4 - \frac{1}{{{4^{2013}}}} \Rightarrow B = \frac{4}{3} - \frac{1}{{{{3.4}^{2013}}}}\\
3A = B - \frac{{2014}}{{{4^{2014}}}} = \frac{4}{3} - \left( {\frac{1}{{{{3.4}^{2013}}}} + \frac{{2014}}{{{4^{2014}}}}} \right) \Rightarrow A = ....
\end{array}\)
