Cho cấp số cộng \(\left( {{u_n}} \right)\) với \({u_1} = 11;\,\,\,{u_2} = 13\). Tính tổng \(S = \dfrac{1}{{{u_1}{u_2}}} + \dfrac{1}{{{u_2}{u_3}}} + .... + \dfrac{1}{{{u_{99}}{u_{10...

* Hướng dẫn giải

Gọi \(d\) là công sai của cấp số cộng ta có: \(d = {u_2} - {u_1} = 13 - 11 = 2\).

Khi đó ta có

\(\begin{array}{l}\dfrac{1}{{{u_n}{u_{n + 1}}}} = \dfrac{1}{{{u_n}\left( {{u_n} + 2} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\left[ {\dfrac{{{u_n} + 2}}{{{u_1}\left( {{u_n} + 2} \right)}} - \dfrac{{{u_n}}}{{{u_1}\left( {{u_n} + 2} \right)}}} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\left[ {\dfrac{1}{{{u_n}}} - \dfrac{1}{{{u_n} + 2}}} \right] = \dfrac{1}{2}\left( {\dfrac{1}{{{u_n}}} - \dfrac{1}{{{u_{n + 1}}}}} \right)\end{array}\)

Suy ra

\(\begin{array}{l}S = \dfrac{1}{{{u_1}{u_2}}} + \dfrac{1}{{{u_2}{u_3}}} + ... + \dfrac{1}{{{u_{99}}{u_{100}}}}\\S = \dfrac{1}{2}\left( {\dfrac{1}{{{u_1}}} - \dfrac{1}{{{u_2}}} + \dfrac{1}{{{u_2}}} - \dfrac{1}{{{u_3}}} + ... + \dfrac{1}{{{u_{99}}}} - \dfrac{1}{{{u_{100}}}}} \right)\\S = \dfrac{1}{2}\left( {\dfrac{1}{{{u_1}}} - \dfrac{1}{{{u_{100}}}}} \right)\\S = \dfrac{1}{2}\left( {\dfrac{1}{{{u_1}}} - \dfrac{1}{{{u_1} + 99d}}} \right)\\S = \dfrac{1}{2}\left( {\dfrac{1}{{11}} - \dfrac{1}{{11 + 99.2}}} \right) = \dfrac{9}{{209}}\end{array}\)

Chọn A.