Giải thích các bước giải:
$A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{399}{400}\\
=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+1-\dfrac{1}{5^2}+...+1-\dfrac{1}{20^2}\\
=(1+1+1+1...+1)-\left( \dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{20^2} \right )\\
=19-\left( \dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{20^2} \right )\\
\dfrac{1}{2^2}<\dfrac{1}{1.2}\\
\dfrac{1}{3^2}<\dfrac{1}{2.3}\\
...\\
\dfrac{1}{20^2}<\dfrac{1}{19.20}\\
\Rightarrow \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{19.20}\\
=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\\
=1-\dfrac{1}{20}=\dfrac{19}{20}<1\\
\Rightarrow A=19-\left ( \dfrac{1}{2^2}+...+\dfrac{1}{20^2} \right )\\
>19-1=18\\
\Rightarrow A>18$
