Cho \(A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots + \frac{1}{{101.102}};\,\,\,B = \frac{1}{{52.102}} + \frac{1}{{53.101}} + \frac{1}{{54.100}} + \ldots + \frac{1}{{102.52}} + \frac{2}{{77.154}}\). Tính \(\frac{A}{B} \)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} A = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots + \frac{1}{{101.102}}\\ \,\,\,\, = \left( {\frac{1}{1} - \frac{1}{2}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \ldots + \frac{1}{{101}} - \frac{1}{{102}} = \left( {\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{{101}} - \frac{1}{{102}}} \right)\\ \,\,\,\, = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{101}} + \frac{1}{{102}}} \right) - 2\left( {\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{{102}}} \right)\\ \,\,\,\, = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{102}}} \right) - \left( {\frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{{51}}} \right) = \frac{1}{{52}} + \frac{1}{{53}} + \ldots + \frac{1}{{101}} + \frac{1}{{102}}\\ \,\,\,\, = \left( {\frac{1}{{52}} + \frac{1}{{102}}} \right) + \left( {\frac{1}{{53}} + \frac{1}{{101}}} \right) + \ldots + \left( {\frac{1}{{76}} + \frac{1}{{78}}} \right) + \frac{1}{{77}} = \frac{{154}}{{52.102}} + \frac{{154}}{{53.101}} + \ldots + \frac{{154}}{{76.78}} + \frac{{154}}{{77.154}}\\ B = \left( {\frac{1}{{52.102}} + \frac{1}{{102.52}}} \right) + \left( {\frac{1}{{53.101}} + \frac{1}{{101.53}}} \right) + \ldots + \left( {\frac{1}{{76.78}} + \frac{1}{{78.76}}} \right) + \frac{2}{{77.154}}\\ \,\,\,\, = \frac{2}{{52.102}} + \frac{2}{{53.101}} + \ldots + \frac{2}{{76.78}} + \frac{2}{{77.154}}\\ \Rightarrow \frac{A}{B} = \frac{{154}}{2} = 77 \end{array}\)
