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Câu hỏi: Cho khai triển ${{\left( 1+x+{{x}^{2}} \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+....+{{a}_{2n}}{{x}^{2n}}$, với $n\ge 2$. Biết rằng $\dfrac{{{a}_{3}}}{14}=\dfrac{{{a}_{4}}}{41}$, khi đó tổng $S={{a}_{0}}+{{a}_{1}}+{{a}_{2}}+...+{{a}_{2n}}$ bằng
A. $S={{3}^{10}}$
B. $S={{3}^{11}}$
C. $S={{3}^{12}}$
D. $S={{3}^{13}}$
A. $S={{3}^{10}}$
B. $S={{3}^{11}}$
C. $S={{3}^{12}}$
D. $S={{3}^{13}}$
Ta có ${{\left( 1+x+{{x}^{2}} \right)}^{n}}=\sum\limits_{k=0}^{n}{C_{n}^{k}{{\left( x+{{x}^{2}} \right)}^{k}}=\sum\limits_{k=0}^{n}{C_{n}^{k}}}\sum\limits_{l=0}^{k}{C_{k}^{l}{{x}^{k-l}}.{{x}^{2l}}}$
Hệ số của ${{x}^{3}}$ là ${{x}^{k+l}}={{x}^{3}}\Rightarrow k+l=3\Rightarrow \left[ \begin{aligned}
& l=0;k=3 \\
& l=1;k=2 \\
\end{aligned} \right.\Rightarrow {{a}_{3}}=C_{n}^{3}C_{3}^{0}+C_{n}^{2}C_{2}^{1}$
Tương tự hệ số của ${{x}^{4}}$ là ${{x}^{k+l}}={{x}^{4}}\Rightarrow k+l=4\Rightarrow \left[ \begin{aligned}
& l=0;k=4 \\
& l=1;k=3 \\
& l=2;k=2 \\
\end{aligned} \right.\Rightarrow {{a}_{4}}=C_{n}^{4}C_{4}^{0}+C_{n}^{3}C_{3}^{1}+C_{n}^{2}C_{2}^{2}$.
Theo giả thiết $14{{a}_{4}}=41{{a}_{3}}\Leftrightarrow 14\left( C_{n}^{4}C_{4}^{0}+C_{n}^{3}C_{3}^{1}+C_{n}^{2}C_{2}^{2} \right)=41\left( C_{n}^{3}C_{3}^{0}+C_{n}^{2}C_{2}^{1} \right)$
$\begin{aligned}
& \Leftrightarrow 14\left( \dfrac{n!}{4!\left( n-4 \right)!}+\dfrac{3.n!}{3!\left( n-3 \right)!}+\dfrac{n!}{2!\left( n-2 \right)!} \right)=41\left( \dfrac{n!}{3!\left( n-3 \right)!}+\dfrac{2.n!}{2!\left( n-2 \right)!} \right) \\
& \Leftrightarrow 14\left( \dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)}{24}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{2}+\dfrac{n\left( n-1 \right)}{2} \right)=41\left( \dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}+n\left( n-1 \right) \right) \\
& \Leftrightarrow n\left( n-1 \right)\left( \dfrac{14}{24}{{n}^{2}}-\dfrac{11}{4}n-\dfrac{185}{6} \right)=0\Leftrightarrow \left[ \begin{aligned}
& n=1 \\
& n=10 \\
\end{aligned} \right.\left( n\in {{\mathbb{N}}^{*}} \right) \\
\end{aligned}$
Do $n\ge 2$ nên $n=10$
Thay $x=1$ vào hai vế của khai triển ${{\left( 1+x+{{x}^{2}} \right)}^{10}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+....+{{a}_{20}}{{x}^{20}}$ ta được
$S={{a}_{0}}+{{a}_{1}}+{{a}_{2}}+....+{{a}_{20}}={{3}^{10}}$
Hệ số của ${{x}^{3}}$ là ${{x}^{k+l}}={{x}^{3}}\Rightarrow k+l=3\Rightarrow \left[ \begin{aligned}
& l=0;k=3 \\
& l=1;k=2 \\
\end{aligned} \right.\Rightarrow {{a}_{3}}=C_{n}^{3}C_{3}^{0}+C_{n}^{2}C_{2}^{1}$
Tương tự hệ số của ${{x}^{4}}$ là ${{x}^{k+l}}={{x}^{4}}\Rightarrow k+l=4\Rightarrow \left[ \begin{aligned}
& l=0;k=4 \\
& l=1;k=3 \\
& l=2;k=2 \\
\end{aligned} \right.\Rightarrow {{a}_{4}}=C_{n}^{4}C_{4}^{0}+C_{n}^{3}C_{3}^{1}+C_{n}^{2}C_{2}^{2}$.
Theo giả thiết $14{{a}_{4}}=41{{a}_{3}}\Leftrightarrow 14\left( C_{n}^{4}C_{4}^{0}+C_{n}^{3}C_{3}^{1}+C_{n}^{2}C_{2}^{2} \right)=41\left( C_{n}^{3}C_{3}^{0}+C_{n}^{2}C_{2}^{1} \right)$
$\begin{aligned}
& \Leftrightarrow 14\left( \dfrac{n!}{4!\left( n-4 \right)!}+\dfrac{3.n!}{3!\left( n-3 \right)!}+\dfrac{n!}{2!\left( n-2 \right)!} \right)=41\left( \dfrac{n!}{3!\left( n-3 \right)!}+\dfrac{2.n!}{2!\left( n-2 \right)!} \right) \\
& \Leftrightarrow 14\left( \dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)}{24}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{2}+\dfrac{n\left( n-1 \right)}{2} \right)=41\left( \dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}+n\left( n-1 \right) \right) \\
& \Leftrightarrow n\left( n-1 \right)\left( \dfrac{14}{24}{{n}^{2}}-\dfrac{11}{4}n-\dfrac{185}{6} \right)=0\Leftrightarrow \left[ \begin{aligned}
& n=1 \\
& n=10 \\
\end{aligned} \right.\left( n\in {{\mathbb{N}}^{*}} \right) \\
\end{aligned}$
Do $n\ge 2$ nên $n=10$
Thay $x=1$ vào hai vế của khai triển ${{\left( 1+x+{{x}^{2}} \right)}^{10}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+....+{{a}_{20}}{{x}^{20}}$ ta được
$S={{a}_{0}}+{{a}_{1}}+{{a}_{2}}+....+{{a}_{20}}={{3}^{10}}$
Đáp án A.