Giải thích các bước giải:
$A=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots+\dfrac{1}{160^2}\\ =\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{80^2}\right)\\ <\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\cdots+\dfrac{1}{79.80}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots+\dfrac{1}{79}-\dfrac{1}{80}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{80}\right)\\ =\dfrac{59}{320}\\ <\dfrac{60}{320}\\ =\dfrac{3}{16}\\ A=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{80^2}\right)\\ >\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\cdots+\dfrac{1}{80.81}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdots+\dfrac{1}{80}-\dfrac{1}{81}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{81}\right)\\ =\dfrac{185}{1296}\\ >\dfrac{162}{1296}\\ =\dfrac{1}{8}\\ \Rightarrow \dfrac{1}{8}<A<\dfrac{3}{16}$
