Tính Lim {1 / {u_1}.{u_2}}} + {1} / {u_2}.{u_3} + ....+ {1}/{u_{n - 1}}.{u_n}

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Cho cấp số cộng \(\left\{ {\begin{array}{*{20}{l}}{{u_1} = 2}\\{{u_n} = {u_{n - 1}} + d}\end{array}} \right.\), (d là công sai)

Tính \({\rm{lim}}\frac{1}{{{u_1}.{u_2}}} + \frac{1}{{{u_2}.{u_3}}} + \ldots + \frac{1}{{{u_{n - 1}}.{u_n}}}\)

\(\frac{1}{6}\).

\( + \infty \).

Giải thích

Đáp án B

Hướng dẫn giải

Ta có :

\(\frac{1}{{{u_1}.{u_2}}} + \frac{1}{{{u_2}.{u_3}}} + \ldots + \frac{1}{{{u_{n - 1}}.{u_n}}} = \frac{1}{{2.5}} + \frac{1}{{5.8}} + \ldots + \frac{1}{{{u_{n - 1}}\left( {{u_{n - 1}} + 3} \right)}}\)

\( = \frac{1}{3}\left( {\frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \ldots + \frac{1}{{{u_{n - 1}}}} - \frac{1}{{{u_{n - 1}} + 3}}} \right) = \frac{1}{3}\left( {\frac{1}{2} - \frac{1}{{2 + \left( {n - 1} \right).3}}} \right)\)

\( \Rightarrow {\rm{lim}}\left( {\frac{1}{{{u_1}.{u_2}}} + \frac{1}{{{u_2}.{u_3}}} + \ldots + \frac{1}{{{u_{n - 1}}.{u_n}}}} \right)\)

\( = {\rm{lim}}\left( {\frac{1}{3}\left( {\frac{1}{2} - \frac{1}{{2 + \left( {n - 1} \right).3}}} \right)} \right) = \frac{1}{6}\)