Cho tổng: S n = 1/ 1.2.3 + 1/ 2.3.4 + 1 /3.4.5 + . . . + 1/ n ( n + 1 ) ( n + 2 ) . Tính S 30

Ta có \(2{S_n} = \frac{2}{{1.2.3}} + \frac{2}{{2.3.4}} + \frac{2}{{3.4.5}} + ... + \frac{2}{{n\left( {n + 1} \right)\left( {n + 2} \right)}}\)

Trong đó

\(\begin{array}{l}\frac{2}{{1.2.3}} = \frac{1}{{1.2}} - \frac{1}{{2.3}};\,\frac{2}{{2.3.4}} = \frac{1}{{2.3}} - \frac{1}{{3.4}};\,\,\frac{2}{{3.4.5}} = \frac{1}{{3.4}} - \frac{1}{{4.5}};\,\,\\\frac{2}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{{n\left( {n + 1} \right)}} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\end{array}\)

Khi đó

\(\begin{array}{l}2{S_n} = \left( {\frac{1}{{1.2}} - \frac{1}{{2.3}}} \right) + \left( {\frac{1}{{2.3}} - \frac{1}{{3.4}}} \right) + \left( {\frac{1}{{3.4}} - \frac{1}{{4.5}}} \right) + ... + \left( {\frac{1}{{n\left( {n + 1} \right)}} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}} \right)\\\,\,\,\,\,\,\, = \frac{1}{{1.2}} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{{{n^2} + 3n}}{{\left( {n + 1} \right)\left( {n + 2} \right)}} \Rightarrow {S_n} = \frac{{{n^2} + 3n}}{{2\left( {n + 1} \right)\left( {n + 2} \right)}}\end{array}\)

Vậy \({S_{30}} = \frac{{{{30}^2} + 3.30}}{{2.\left( {30 + 1} \right)\left( {30 + 2} \right)}} = \frac{{495}}{{992}}\)