Đáp án:
$\begin{array}{l}
A = \frac{3}{4} + \frac{8}{9} + \frac{{15}}{{16}} + ... + \frac{{399}}{{400}}\\
= \left( {1 - \frac{1}{4}} \right) + \left( {1 - \frac{1}{9}} \right) + \left( {1 - \frac{1}{{16}}} \right) + ... + \left( {1 - \frac{1}{{400}}} \right)\\
= \left( {1 + 1 + ... + 1} \right) - \left( {\frac{1}{4} + \frac{1}{9} + \frac{1}{{16}} + ... + \frac{1}{{400}}} \right)\\
= S - \left( {\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{20}^2}}}} \right)\\
= S - B
\end{array}$
Tổng S có: (20-2):1+1=19 số hạng
=> S=19
$\begin{array}{l}
B = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{20}^2}}}\\
Do:\frac{1}{{{2^2}}} < \frac{1}{{1.2}}\\
\frac{1}{{{3^2}}} < \frac{1}{{2.3}}\\
\frac{1}{{{4^2}}} < \frac{1}{{3.4}}\\
...\frac{1}{{{{20}^2}}} < \frac{1}{{19.20}}\\
\Rightarrow B < \frac{1}{{1.2}} + \frac{1}{{2.3}} + ... + \frac{1}{{19.20}}\\
\Rightarrow B < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{{19}} - \frac{1}{{20}}\\
\Rightarrow B < 1 - \frac{1}{{20}} < 1\\
\Rightarrow S - B > 19 - 1\\
\Rightarrow A > 18
\end{array}$
