Đáp án:
Câu này chỗ +1/5.3 phải là -1/5.3 mới tính được nhé!
$\begin{array}{l}
F = \dfrac{1}{{99}} - \dfrac{1}{{99.97}} - \dfrac{1}{{97.95}} - .... - \dfrac{1}{{5.3}} - \dfrac{1}{{3.1}}\\
= \dfrac{1}{{99}} - \left( {\dfrac{1}{{99.97}} + \dfrac{1}{{97.95}} + .... + \dfrac{1}{{5.3}} + \dfrac{1}{{3.1}}} \right)\\
= \dfrac{1}{{99}} - \dfrac{1}{2}.\left( {\dfrac{2}{{99.97}} + \dfrac{2}{{97.95}} + ... + \dfrac{2}{{5.3}} + \dfrac{2}{{3.1}}} \right)\\
= \dfrac{1}{{99}} - \dfrac{1}{2}.\left( {\dfrac{1}{{97}} - \dfrac{1}{{99}} + \dfrac{1}{{95}} - \dfrac{1}{{97}} + ... + \dfrac{1}{3} - \dfrac{1}{5} + 1 - \dfrac{1}{3}} \right)\\
= \dfrac{1}{{99}} - \dfrac{1}{2}.\left( {1 - \dfrac{1}{{99}}} \right)\\
= \dfrac{1}{{99}} - \dfrac{1}{2}.\dfrac{{98}}{{99}}\\
= \dfrac{1}{{99}} - \dfrac{{49}}{{99}}\\
= \dfrac{{ - 48}}{{99}}
\end{array}$
