Giải thích các bước giải:
+\[\begin{array}{l}
A = \dfrac{1}{{{2^2}{{.2}^2}}} + \dfrac{1}{{{2^2}{{.3}^2}}} + \dfrac{1}{{{2^2}{{.4}^2}}} + ... + \dfrac{1}{{{2^2}{{.80}^2}}}\\
\Rightarrow {2^2}A = {2^2}.\left( {\dfrac{1}{{{2^2}{{.2}^2}}} + \dfrac{1}{{{2^2}{{.3}^2}}} + \dfrac{1}{{{2^2}{{.4}^2}}} + ... + \dfrac{1}{{{2^2}{{.80}^2}}}} \right)\\
\Rightarrow 4A = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + ... + \dfrac{1}{{{{80}^2}}}\\
\dfrac{1}{{{3^2}}} < \dfrac{1}{{2.3}} = \dfrac{1}{2} - \dfrac{1}{3}\\
\dfrac{1}{{{4^2}}} < \dfrac{1}{{3.4}} = \dfrac{1}{3} - \dfrac{1}{4}\\
....\\
\dfrac{1}{{{{80}^2}}} < \dfrac{1}{{79.80}} < \dfrac{1}{{79}} - \dfrac{1}{{80}}\\
\Rightarrow 4A < \dfrac{1}{{{2^2}}} + \dfrac{1}{2} - \dfrac{1}{{80}} = \dfrac{{59}}{{80}} \Rightarrow A < \dfrac{{59}}{{320}} < \dfrac{{60}}{{320}} = \dfrac{3}{{16}}
\end{array}\]
Vậy $A<\dfrac{3}{16}$
+ Chứng minh $A>\dfrac{1}{8}$
\[\begin{array}{l}
A = \dfrac{1}{{{2^2}{{.2}^2}}} + \dfrac{1}{{{2^2}{{.3}^2}}} + \dfrac{1}{{{2^2}{{.4}^2}}} + ... + \dfrac{1}{{{2^2}{{.80}^2}}}\\
\Rightarrow {2^2}A = {2^2}.\left( {\dfrac{1}{{{2^2}{{.2}^2}}} + \dfrac{1}{{{2^2}{{.3}^2}}} + \dfrac{1}{{{2^2}{{.4}^2}}} + ... + \dfrac{1}{{{2^2}{{.80}^2}}}} \right)\\
\Rightarrow 4A = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + ... + \dfrac{1}{{{{80}^2}}}\\
\dfrac{1}{{{3^2}}} > \dfrac{1}{{3.4}} = \dfrac{1}{3} - \dfrac{1}{4}\\
\dfrac{1}{{{4^2}}} > \dfrac{1}{{4.5}} = \dfrac{1}{4} - \dfrac{1}{5}\\
....\\
\dfrac{1}{{{{80}^2}}} > \dfrac{1}{{80.81}} < \dfrac{1}{{80}} - \dfrac{1}{{81}}\\
\Rightarrow 4A > \dfrac{1}{{{2^2}}} + \dfrac{1}{3} - \dfrac{1}{{81}} = \dfrac{{185}}{{324}} \Rightarrow A > \dfrac{{185}}{{1296}} > \dfrac{{162}}{{1296}} = \dfrac{1}{8}
\end{array}\]
Vậy $A>\dfrac{1}{8}$
