Cho \(A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots + \frac{1}{{199.200}};\,\,B = \frac{1}{{101.200}} + \frac{1}{{102.199}} + \ldots + \frac{1}{{200.101}}\\\). Tính \(\frac{A}{B} \)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots + \frac{1}{{199.200}}\\ \,\,\,\, = \left( {\frac{1}{1} - \frac{1}{2}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \ldots + \left( {\frac{1}{{199}} - \frac{1}{{200}}} \right) = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{200}}} \right) - 2\left( {\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{{200}}} \right)\\ \,\,\,\, = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{200}}} \right) - \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{100}}} \right) = \frac{1}{{101}} + \frac{1}{{102}} + \ldots + \frac{1}{{200}}\\ \,\,\,\, = \left( {\frac{1}{{101}} + \frac{1}{{200}}} \right) + \left( {\frac{1}{{102}} + \frac{1}{{199}}} \right) + \ldots + \left( {\frac{1}{{150}} + \frac{1}{{151}}} \right) = \frac{{301}}{{101.200}} + \frac{{301}}{{102.199}} + \ldots + \frac{{301}}{{150.151}}\\ B = \left( {\frac{1}{{101.200}} + \frac{1}{{200.101}}} \right) + \left( {\frac{1}{{102.199}} + \frac{1}{{199.102}}} \right) + \ldots + \left( {\frac{1}{{150.151}} + \frac{1}{{151.150}}} \right)\\ \,\,\,\, = \frac{2}{{101.200}} + \frac{2}{{102.199}} + \ldots + \frac{2}{{150.151}}\\ \Rightarrow \frac{A}{B} = \frac{{301}}{2} \end{array}\)
