Với \(n \in {\mathbb{N}^*}\), tính tổng sau: \(A = \frac{1}{{1.3}} + \frac{1}{{3.5}}
Giải thích
Ta có: \(A = \frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + ... + \frac{1}{{\left( {2n - 1} \right)\left( {2x + 1} \right)}}\)
\(A = \frac{1}{2}\left[ {\frac{2}{{1.3}} + \frac{2}{{3.5}} + \frac{2}{{5.7}} + ... + \frac{2}{{\left( {2n - 1} \right)\left( {2x + 1} \right)}}} \right]\)
\(A = \frac{1}{2}\left[ {1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + ... + \frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right]\)
\(A = \frac{1}{2}\left( {1 - \frac{1}{{2n + 1}}} \right)\)
\(A = \frac{1}{2}.\frac{{2n}}{{2n + 1}} = \frac{n}{{2n + 1}}\).
Vậy \(A = \frac{n}{{2n + 1}}\).