Với \(n \in {\mathbb{N}^*}\), hãy tính tổng \(B = \frac{1}{{1.5}} + \frac{1}{{5.9}} + ... + \frac{1}{{\left( {4n - 3} \right)\left( {4n + 1} \right)}}\).
Giải thích
Ta có: \(B = \frac{1}{{1.5}} + \frac{1}{{5.9}} + ... + \frac{1}{{\left( {4n - 3} \right)\left( {4n + 1} \right)}}\)
\(B = \frac{1}{4}\left[ {\frac{4}{{1.5}} + \frac{4}{{5.9}} + ... + \frac{4}{{\left( {4n - 3} \right)\left( {4n + 1} \right)}}} \right]\)
\(B = \frac{1}{4}\left[ {1 - \frac{1}{5} + \frac{1}{5} - \frac{1}{9} + ... + \frac{1}{{\left( {4n - 3} \right)}} - \frac{1}{{\left( {4n + 1} \right)}}} \right]\)
\(B = \frac{1}{4}\left[ {1 - \frac{1}{{\left( {4n + 1} \right)}}} \right]\)
\(B = \frac{1}{4}.\frac{{4n}}{{\left( {4n + 1} \right)}}\)
\(B = \frac{n}{{\left( {4n + 1} \right)}}\).