Cho \(A = 1 + \frac{1}{3} + \frac{1}{5} + \ldots + \frac{1}{{999}};B = \frac{1}{{1.999}} + \frac{1}{{3.997}} + \frac{1}{{5.1995}} + \ldots + \frac{1}{{999.1}}\). Tính \(\frac{A}{B}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} A = \left( {1 + \frac{1}{{999}}} \right) + \left( {\frac{1}{3} + \frac{1}{{997}}} \right) + \ldots + \left( {\frac{1}{{499}} + \frac{1}{{501}}} \right) = \frac{{1000}}{{999.1}} + \frac{{1000}}{{3.997}} + \ldots + \frac{{1000}}{{499.501}}\\ \,\,\,\, = 1000\left( {\frac{1}{{999.1}} + \frac{1}{{3.997}} + .. + \frac{1}{{499.501}}} \right)\\ B = \left( {\frac{1}{{1.999}} + \frac{1}{{999.1}}} \right) + \left( {\frac{1}{{3.997}} + \frac{1}{{997.3}}} \right) + \ldots + \left( {\frac{1}{{499.501}} + \frac{1}{{501.499}}} \right)\\ \,\,\,\, = \frac{2}{{1.999}} + \frac{2}{{3.997}} + \ldots + \frac{2}{{499.501}}\\ \,\,\,\, = 2\left( {\frac{1}{{1.999}} + \frac{1}{{3.997}} + \ldots + \frac{1}{{499.501}}} \right)\\ \Rightarrow \frac{A}{B} = \frac{{1000}}{2} = 500 \end{array}\)
