Cho \(E = \frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + \ldots + \frac{1}{{100}};F = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots + \frac{1}{{99.100}}\). Khẳng định nào sau đây đúng?

Ta có:

\(\begin{array}{l} F = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots + \frac{1}{{99.100}}\\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots \ldots + \frac{1}{{99}} - \frac{1}{{100}}\\ = \left( {\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \ldots \ldots . + \frac{1}{{99}}} \right) - \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots \ldots + \frac{1}{{100}}} \right)\\ = \left( {\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \ldots \ldots . + \frac{1}{{99}}} \right) + \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots \ldots + \frac{1}{{100}}} \right)\\ - \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots \ldots + \frac{1}{{100}}} \right) - \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots \ldots + \frac{1}{{100}}} \right)\\ = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \ldots \ldots + \frac{1}{{99}} + \frac{1}{{100}} - 2 \cdot \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots .. + \frac{1}{{100}}} \right)\\ = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \ldots \ldots + \frac{1}{{49}} + \frac{1}{{50}}} \right) + \frac{1}{{51}} + \frac{1}{{52}} \ldots .. + \frac{1}{{99}} + \frac{1}{{100}} - \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots \ldots + \frac{1}{{50}}} \right)\\ = \frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + \ldots + \frac{1}{{100}} = E\\ \Rightarrow E = F \end{array}\)