Cho \(A = \frac{{2012}}{{51}} + \frac{{2012}}{{52}} + \frac{{2012}}{{53}} + \ldots + \frac{{2012}}{{100}};\,\,B = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots + \frac{1}{{99.100}}\). Tính \(\frac{A}{B} \).
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Lời giải:
Báo saiTa có
\(\begin{array}{l} A = 2012\left( {\frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + \ldots + \frac{1}{{100}}} \right)\\ B = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{{99}} - \frac{1}{{100}} = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{99}} + \frac{1}{{100}}} \right) - 2\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{{100}}} \right)\\ \,\,\, = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{100}}} \right) - \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{50}}} \right) = \frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + \ldots + \frac{1}{{100}}\\ \Rightarrow \frac{A}{B} = \frac{{2012}}{1} = 2012 \end{array}\)
