Tính: \(S = 1 + + 2}} + + 2 + 3}} + + 2 + 3 + 4}} + \) \(... + + 2 + 3 + 4 + ... + 8}}\)

* Hướng dẫn giải

\(S = 1 + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + \dfrac{1}{{1 + 2 + 3 + 4}}\) \( + ... + \dfrac{1}{{1 + 2 + 3 + 4 + ... + 8}}\)

\(S = 1 + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + \dfrac{1}{{1 + 2 + 3 + 4}}\)\( + ... + \dfrac{1}{{1 + 2 + 3 + 4 + ... + 8}}\)

\( = 1 + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{{10}} + ... + \dfrac{1}{{36}}\)

\( \Rightarrow \dfrac{1}{2}.S = \dfrac{1}{2}\left( {1 + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{{10}} + ... + \dfrac{1}{{36}}} \right)\)

\( = \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{{12}} + \dfrac{1}{{20}} + ... + \dfrac{1}{{72}}\)

\( = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}}\)\( + ... + \dfrac{1}{{8.9}}\)

\(1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4}\)\( + \dfrac{1}{4} - \dfrac{1}{5} + ... + \dfrac{1}{8} - \dfrac{1}{9}\)

\(\begin{array}{l}\, = 1 - \dfrac{1}{9} = \dfrac{8}{9}\\ \Rightarrow \dfrac{1}{2}S = \dfrac{8}{9}\\ \Rightarrow S = \dfrac{8}{9}:\dfrac{1}{2} = \dfrac{{16}}{9}.\end{array}\)