$\left( {1 - \dfrac{1}{{1 + 2}}} \right)\left( {1 - \dfrac{1}{{1 + 2 + 3}}} \right)\left( {1 - \dfrac{1}{{1 + 2 + 3 + 4}}} \right)...\left( {1 - \dfrac{1}{{1 + 2 + 3 + ... + 50}}} \right)$
Nhận xét: $1 - \dfrac{1}{{1 + 2 + 3 + ... + n}} = 1 - \dfrac{2}{{n\left( {n + 1} \right)}} = \dfrac{{{n^2} + n - 2}}{{n\left( {n + 1} \right)}} = \dfrac{{\left( {n - 1} \right)\left( {n + 2} \right)}}{{n\left( {n + 1} \right)}}$
Do đó: $\left( {1 - \dfrac{1}{{1 + 2}}} \right)\left( {1 - \dfrac{1}{{1 + 2 + 3}}} \right)...\left( {1 - \dfrac{1}{{1 + 2 + 3 + ... + 50}}} \right) = \dfrac{{1.4}}{{2.3}}.\dfrac{{2.5}}{{3.4}}...\dfrac{{49.52}}{{50.51}} = \dfrac{{\left( {1.2.3...49} \right).\left( {4.5.6...52} \right)}}{{\left( {2.3....50} \right)\left( {3.4...51} \right)}} = \dfrac{{1.4.52}}{{50.3}} = \dfrac{{104}}{{75}}$
