Đáp án:
\[B = \frac{{ - 2019}}{{4036}}\]
Giải thích các bước giải:
Ta có:
\[\frac{1}{{{n^2}}} - 1 = \frac{{1 - {n^2}}}{{{n^2}}} = - \frac{{{n^2} - 1}}{{{n^2}}} = - \frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}\]
Suy ra:
\(\begin{array}{l}
B = \left( {\frac{1}{{{{2018}^2}}} - 1} \right)\left( {\frac{1}{{{{2017}^2}}} - 1} \right)\left( {\frac{1}{{{{2016}^2}}} - 1} \right).....\left( {\frac{1}{{{2^2}}} - 1} \right)\\
= \left( { - \frac{{2017.2019}}{{{{2018}^2}}}} \right).\left( { - \frac{{2016.2018}}{{{{2017}^2}}}} \right).\left( { - \frac{{2015.2017}}{{{{2016}^2}}}} \right).....\left( { - \frac{{1.3}}{{{2^2}}}} \right)\\
= - \frac{{\left( {2017.2016.2015.....1} \right).\left( {2019.2018.2017.....3} \right)}}{{{{2018}^2}{{.2017}^2}{{.2016}^2}{{.....2}^2}}}\\
= - \frac{{2017.2016.2015....1}}{{2018.2017.2016....2}}.\frac{{2019.2018.2017....3}}{{2018.2017.2016....2}}\\
= - \frac{1}{{2018}}.\frac{{2019}}{2}\\
= \frac{{ - 2019}}{{4036}}
\end{array}\)
