Cho \(E = \left( {1 - \frac{1}{{1 + 2}}} \right)\left( {1 - \frac{1}{{1 + 2 + 3}}} \right) \ldots \left( {1 - \frac{1}{{1 + 2 + 3 + \ldots + n}}} \right)\), \(F = \frac{{n + 2}}{n}\). Tính \(\frac{E}{F}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} E = \left( {1 - \frac{1}{{1 + 2}}} \right)\left( {1 - \frac{1}{{1 + 2 + 3}}} \right) \ldots \left( {1 - \frac{1}{{1 + 2 + 3 + \ldots + n}}} \right)\\ E = \left( {1 - \frac{1}{{\frac{{(1 + 2).2}}{2}}}} \right)\left( {1 - \frac{1}{{\frac{{(1 + 3).3}}{2}}}} \right) \ldots \left( {1 - \frac{1}{{\frac{{(1 + n).n}}{2}}}} \right)\\ = \left( {1 - \frac{2}{{2.3}}} \right)\left( {1 - \frac{2}{{3.4}}} \right)\left( {1 - \frac{2}{{4.5}}} \right) \ldots \left( {1 - \frac{2}{{n(n + 1)}}} \right) = \frac{4}{{2.3}} \cdot \frac{{10}}{{3.4}} \cdot \frac{{18}}{{4.5}} \ldots \frac{{n(n + 1) - 2}}{{n(n + 1)}}\\ = \frac{{1.4}}{{2.3}} \cdot \frac{{2.5}}{{3.4}} \cdot \frac{{3.6}}{{4.5}} \ldots \frac{{(n - 1)(n + 2)}}{{n(n + 1)}} = \frac{{(1.2.3 \ldots ({\rm{n}} - 1))(4.5 \ldots .(n + 2))}}{{(2.3 \ldots n)(3.4.5 \ldots (n + 1))}} = \frac{{n + 2}}{{n.3}} = \frac{{n + 2}}{{3n}}\\ \Rightarrow \frac{E}{F} = \frac{{n + 2}}{{3n}}:\frac{{n + 2}}{n} = \frac{1}{3} \end{array}\)
