Bài tập 4.49 trang 173 SBT Toán 11

Ta có:

\(\begin{array}{l}
2,131131131... = 2 + \frac{{131}}{{{{10}^3}}} + \frac{{131}}{{{{10}^6}}} + \frac{{131}}{{{{10}^9}}} + ...\\
 = 2 + \frac{{131}}{{{{10}^3}}}\left( {1 + \frac{1}{{{{10}^3}}} + \frac{1}{{{{10}^6}}} + \frac{1}{{{{10}^9}}} + ...} \right) = 2 + \frac{{131}}{{{{10}^3}}}.S
\end{array}\)

Ta có:

\(S = 1 + \frac{1}{{{{10}^3}}} + \frac{1}{{{{10}^6}}} + \frac{1}{{{{10}^9}}} + ...\) là cấp số nhân lùi vô hạn có \({u_1} = 1,\,q = \frac{1}{{{{10}^3}}}\) nên 

\(S = \frac{{{u_1}}}{{1 - q}} = \frac{1}{{1 - \frac{1}{{{{10}^3}}}}} = \frac{{1000}}{{999}}\)

Vậy \(2,131131131... = 2 + \frac{{131}}{{{{10}^3}}}.\frac{{1000}}{{999}} = 2 + \frac{{131}}{{999}} = \frac{{2129}}{{999}}\)

-- Mod Toán 11 HỌC247