Lời giải:Với \(n\ge 2,n\in \mathbb{N}*\) ta có:
\(C_{n}^{0}+C_{n}^{1}+C_{n}^{2}=11\Leftrightarrow \frac{n!}{0!.\left( n-0 \right)!}+\frac{n!}{1!.\left( n-1 \right)!}+\frac{n!}{2!.\left( n-2 \right)!}=11\)
\( \Leftrightarrow n + \frac{{n\left( {n - 1} \right)}}{2} = 10 \Leftrightarrow {n^2} + n - 20 = 0 \Leftrightarrow \left[ \begin{array}{l}
n = - 5\\
n = 4
\end{array} \right. \Rightarrow n = 4\)
\(n=4\Rightarrow {{\left( {{x}^{3}}-\frac{1}{{{x}^{2}}} \right)}^{n}}={{\left( {{x}^{3}}-\frac{1}{{{x}^{2}}} \right)}^{4}}\)
\({{\left( {{x}^{3}}-\frac{1}{{{x}^{2}}} \right)}^{4}}=\sum\limits_{k=0}^{4}{C_{4}^{k}.{{\left( {{x}^{3}} \right)}^{4-k}}.\frac{{{\left( -1 \right)}^{k}}}{{{\left( {{x}^{2}} \right)}^{k}}}=}\sum\limits_{k=0}^{4}{{{\left( -1 \right)}^{k}}.C_{4}^{k}.{{x}^{12-5k}}\left( 0\le k\le 4,k\in \mathbb{N} \right)}\)
Số hạng tổng quát \({{\left( -1 \right)}^{k}}C_{4}^{k}.{{x}^{12-5k}}\)
Phải có \({{x}^{12-5k}}={{x}^{7}}\Rightarrow 12-5k=7\Leftrightarrow k=1.\)
Số hạng chứa \({{x}^{7}}\) trong khai triển là: \({{\left( -1 \right)}^{1}}C_{4}^{1}.{{x}^{7}}=-4{{x}^{7}}.\)
