Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{1}{5} + \dfrac{2}{{{5^2}}} + \dfrac{3}{{{5^3}}} + ..... + \dfrac{{2020}}{{{5^{2020}}}}\\
\Rightarrow 5A = 1 + \dfrac{2}{5} + \dfrac{3}{{{5^2}}} + \dfrac{4}{{{5^3}}} + ..... + \dfrac{{2020}}{{{5^{2019}}}}\\
\Rightarrow 5A - A = \left( {1 + \dfrac{2}{5} + \dfrac{3}{{{5^2}}} + \dfrac{4}{{{5^3}}} + ..... + \dfrac{{2020}}{{{5^{2019}}}}} \right) - \left( {\dfrac{1}{5} + \dfrac{2}{{{5^2}}} + \dfrac{3}{{{5^3}}} + ..... + \dfrac{{2020}}{{{5^{2020}}}}} \right)\\
\Leftrightarrow 4A = 1 + \left( {\dfrac{2}{5} - \dfrac{1}{5}} \right) + \left( {\dfrac{3}{{{5^2}}} - \dfrac{2}{{{5^2}}}} \right) + ..... + \left( {\dfrac{{2020}}{{{5^{2019}}}} - \dfrac{{2019}}{{{5^{2019}}}}} \right) - \dfrac{{2020}}{{{5^{2020}}}}\\
\Leftrightarrow 4A = 1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + .... + \dfrac{1}{{{5^{2019}}}} - \dfrac{{2020}}{{{5^{2020}}}}\\
B = \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + .... + \dfrac{1}{{{5^{2019}}}}\\
\Rightarrow 5B = 1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + .... + \dfrac{1}{{{5^{2018}}}}\\
\Rightarrow 5B - B = \left( {1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + .... + \dfrac{1}{{{5^{2018}}}}} \right) - \left( {\dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + .... + \dfrac{1}{{{5^{2019}}}}} \right)\\
\Leftrightarrow 4B = 1 - \dfrac{1}{{{5^{2019}}}}\\
\Rightarrow B = \dfrac{1}{4} - \dfrac{1}{{{{4.5}^{2019}}}}\\
\Rightarrow 4A = 1 + B - \dfrac{{2020}}{{{5^{2020}}}}\\
\Rightarrow 4A = 1 + \dfrac{1}{4} - \dfrac{1}{{{{4.5}^{2019}}}} - \dfrac{{2020}}{{{5^{2020}}}}\\
\Rightarrow A = \dfrac{5}{{16}} - \dfrac{1}{{{4.5^{2019}}}}\left( {\dfrac{1}{4} + \dfrac{{2020}}{5}} \right) = \dfrac{5}{{16}} - \dfrac{{1617}}{{{{16.5}^{2019}}}}\\
\dfrac{1}{{16}} > \dfrac{{1617}}{{{{16.5}^{2019}}}} \Rightarrow A = \dfrac{1}{4} + \left( {\dfrac{1}{{16}} - \dfrac{{1617}}{{{{16.5}^{2019}}}}} \right) > \dfrac{1}{4}\\
\dfrac{5}{{16}} < \dfrac{1}{3} \Rightarrow A < \dfrac{1}{3}\\
\Rightarrow \dfrac{1}{4} < A < \dfrac{1}{3}
\end{array}\)
