Cho \(A = \frac{1}{{1.300}} + \frac{1}{{2.301}} + \ldots + \frac{1}{{101.400}};\,\,B = \frac{1}{{1.102}} + \frac{1}{{2.103}} + \frac{1}{{3.104}} + \ldots + \frac{1}{{299.400}}\). Tính \( \frac{A}{B}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} A = \frac{1}{{1.300}} + \frac{1}{{2.301}} + \ldots + \frac{1}{{101.400}}\\ \Rightarrow 299A = \frac{{299}}{{1.300}} + \frac{{299}}{{2.301}} + \ldots + \frac{{299}}{{101.400}} = \left( {\frac{1}{1} - \frac{1}{{300}}} \right) + \left( {\frac{1}{2} - \frac{1}{{301}}} \right) + \left( {\frac{1}{3} - \frac{1}{{302}}} \right) + \ldots + \left( {\frac{1}{{101}} - \frac{1}{{400}}} \right)\\ \Rightarrow 299A = \left( {1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{101}}} \right) - \left( {\frac{1}{{300}} + \frac{1}{{301}} + \ldots + \frac{1}{{400}}} \right)\\ B = \frac{1}{{1.102}} + \frac{1}{{2.103}} + \frac{1}{{3.104}} + \ldots + \frac{1}{{299.400}}\\ \Rightarrow 101B = \frac{{101}}{{1.102}} + \frac{{101}}{{2.103}} + \frac{{101}}{{3.104}} + \ldots + \frac{{101}}{{299.400}}\\ = \left( {1 - \frac{1}{{102}}} \right) + \left( {\frac{1}{2} - \frac{1}{{103}}} \right) + \left( {\frac{1}{3} - \frac{1}{{104}}} \right) + \ldots + \left( {\frac{1}{{299}} - \frac{1}{{400}}} \right)\\ = \left( {1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{299}}} \right) - \left( {\frac{1}{{102}} + \frac{1}{{103}} + \ldots + \frac{1}{{400}}} \right) = \left( {1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{101}}} \right) - \left( {\frac{1}{{300}} + \frac{1}{{301}} + \ldots + \frac{1}{{400}}} \right)\\ \Rightarrow 299A = 101B \Rightarrow \frac{A}{B} = \frac{{101}}{{299}} \end{array}\)