Cho \(A = 92 - \frac{1}{9} - \frac{2}{{10}} - \frac{3}{{11}} - \ldots - \frac{{92}}{{100}};B = \frac{1}{{45}} + \frac{1}{{50}} + \frac{1}{{55}} + \ldots + \frac{1}{{500}}\). Tính \( \frac{A}{B}\)
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Lời giải:
Báo sai\(\begin{array}{l} A = 92 - \frac{1}{9} - \frac{2}{{10}} - \frac{3}{{11}} - \ldots - \frac{{92}}{{100}}\\ = \left( {1 - \frac{1}{9}} \right) + \left( {1 - \frac{2}{{10}}} \right) + \left( {1 - \frac{3}{{11}}} \right) + \ldots + \left( {1 - \frac{{92}}{{100}}} \right) = \frac{8}{9} + \frac{8}{{10}} + \ldots + \frac{8}{{100}} = 8\left( {\frac{1}{9} + \frac{1}{{10}} + \ldots + \frac{1}{{100}}} \right)\\ B = \frac{1}{5}\left( {\frac{1}{9} + \frac{1}{{10}} + \ldots + \frac{1}{{100}}} \right)\\ \Rightarrow \frac{A}{B} = \frac{8}{{\frac{1}{5}}} = 40 \end{array}\)