Cho \[A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + .... + \frac{1}{{99.100}}\]

                       \(B = \frac{1}{{51.100}} + \frac{1}{{52.99}} + .... + \frac{1}{{99.52}} + \frac{1}{{100.51}}\)

     Tính: \(\frac{A}{B}\).

\[\begin{array}{l}A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + .... + \frac{1}{{99.100}}\\ = \frac{{2 - 1}}{{1.2}} + \frac{{4 - 3}}{{3.4}} + \frac{{6 - 5}}{{5.6}} + .... + \frac{{100 - 99}}{{99.100}}\\ = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + .... + \frac{1}{{99}} - \frac{1}{{100}}\\ = \left( {1 + \frac{1}{3} + \frac{1}{5} + .... + \frac{1}{{99}}} \right) - \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + .... + \frac{1}{{100}}} \right)\\ = \left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + .... + \frac{1}{{100}}} \right) - 2\left( {1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{{50}}} \right)\\ = \frac{1}{{51}} + \frac{1}{{52}} + .... + \frac{1}{{100}}\end{array}\]

Mặt khác:

\[\begin{array}{l}151B = \frac{{51 + 100}}{{51.100}} + \frac{{52 + 99}}{{52.99}} + .... + \frac{{99 + 52}}{{99.52}} + \frac{{100 + 51}}{{100.51}}\\ = \frac{1}{{100}} + \frac{1}{{51}} + \frac{1}{{99}} + \frac{1}{{52}} + .... + \frac{1}{{52}} + \frac{1}{{99}} + \frac{1}{{51}} + \frac{1}{{100}}\\ = \left( {\frac{1}{{100}} + \frac{1}{{99}} + .... + \frac{1}{{52}} + \frac{1}{{51}}} \right) + \left( {\frac{1}{{51}} + \frac{1}{{52}} + .... + \frac{1}{{99}} + \frac{1}{{100}}} \right)\\ = 2\left( {\frac{1}{{51}} + \frac{1}{{52}} + .... + \frac{1}{{99}} + \frac{1}{{100}}} \right)\\ = 2A\\ \Rightarrow \frac{A}{B} = \frac{{151}}{2}\end{array}\]

Vậy \[\frac{A}{B} = \frac{{151}}{2}\].