Cho \(D = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \ldots .\frac{{99}}{{100}}\). Khẳng định nào sau đây đúng?
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Lời giải:
Báo sai\(\begin{array}{l} {\rm{D}} > \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} \ldots \ldots \cdot \frac{{98}}{{99}}\\ \Rightarrow {{\rm{D}}^2} > \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \ldots .\frac{{98}}{{99}} \cdot \frac{{99}}{{100}}\\ \Rightarrow {{\rm{D}}^2} > \frac{1}{{200}}\,mà\,\frac{1}{{200}} > \frac{1}{{225}} \Rightarrow {{\rm{D}}^2} > \frac{1}{{225}} \Rightarrow {\rm{D}} > \frac{1}{{15}}\left( 1 \right)\\ Lại\,có\,D < \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \ldots \cdot \frac{{100}}{{101}}\\ \Rightarrow {{\rm{D}}^2} < \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \ldots ..\frac{{99}}{{100}} \cdot \frac{{100}}{{101}}\\ \Rightarrow {{\rm{D}}^2} < \frac{1}{{101}}ma\frac{1}{{101}} < \frac{1}{{100}} \Rightarrow {{\rm{D}}^2} < \frac{1}{{100}} \Rightarrow {\rm{D}} < \frac{1}{{10}}\left( 2 \right)\\ \left( 1 \right)\left( 2 \right) \Rightarrow \frac{1}{{15}} < D < \frac{1}{{10}} \end{array}\)