Tìm số hạng chứa \(x^3\) trong khai triển \({(x -...

A

- Hướng dẫn giải

\(\begin{array}{l}
\frac{1}{{A_2^2}} + \frac{1}{{A_3^2}} + \frac{1}{{A_4^2}} + ... + \frac{1}{{A_n^2}} = \frac{8}{9}\\
 \Leftrightarrow \frac{{\left( {2 - 2} \right)!}}{{2!}} + \frac{{\left( {3 - 2} \right)!}}{{3!}} + \frac{{\left( {4 - 2} \right)!}}{{4!}} + ... + \frac{{\left( {n - 2} \right)!}}{{n!}} = \frac{8}{9}\\
 \Leftrightarrow 1 + \frac{1}{6} + \frac{1}{{12}} + ... + \frac{1}{{n\left( {n - 1} \right)}} = \frac{8}{9}\\
 \Leftrightarrow 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{n - 1}} - \frac{1}{n} = \frac{8}{9}\\
 \Leftrightarrow 1 - \frac{1}{n} = \frac{8}{9}\\
 \Leftrightarrow n = 9
\end{array}\)

\(\begin{array}{l}
{\left( {x - \frac{2}{{{x^2}}}} \right)^9} = \sum\limits_{k = 0}^9 {C_9^k.{x^{9 - k}}.{{\left( { - \frac{2}{{{x^2}}}} \right)}^k}} \\
 = \sum\limits_{k = 0}^9 {C_9^k.{{\left( { - 2} \right)}^k}.{x^{9 - 3k}}} 
\end{array}\)

Ta có: \(9 - 3k = 3 \Leftrightarrow k = 2\)

Suy ra số hạng chứa \(x^3\) là:

\(C_9^2.{\left( { - 2} \right)^2}{x^3} = 144{x^3}\)