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Câu hỏi: Cho khai triển ${{\left( 1+x+{{x}^{2}} \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+\ldots +{{a}_{2n}}{{x}^{2n}}$, với $n\ge 2$ và ${{a}_{0}}$, ${{a}_{1}}$, ${{a}_{2}}$, ..., ${{a}_{2n}}$ là các hệ số. Biết rằng $\dfrac{{{a}_{3}}}{14}=\dfrac{{{a}_{4}}}{41}$, khi đó tổng $S={{a}_{0}}+{{a}_{1}}+{{a}_{2}}+\ldots +{{a}_{2n}}$ bằng
A. $S={{3}^{11}}$.
B. $S={{3}^{13}}$.
C. $S={{3}^{10}}$.
D. $S={{3}^{12}}$.
A. $S={{3}^{11}}$.
B. $S={{3}^{13}}$.
C. $S={{3}^{10}}$.
D. $S={{3}^{12}}$.
Đặt $f(x)={{\left( 1+x+{{x}^{2}} \right)}^{n}}$.
Khi đó:
$f(x)={{\left( 1+x+{{x}^{2}} \right)}^{n}}={{\left( x+{{x}^{2}}+1 \right)}^{n}}=\sum\limits_{k=0}^{n}{C_{n}^{k}}{{(x+{{x}^{2}})}^{k}}{{.1}^{n-k}}=\sum\limits_{k=0}^{n}{C_{n}^{k}}{{(x+{{x}^{2}})}^{k}}=\sum\limits_{k=0}^{n}{C_{n}^{k}}{{x}^{k}}{{(x+1)}^{k}}=$
$=\sum\limits_{k=0}^{n}{C_{n}^{k}{{x}^{k}}}\left[ \sum\limits_{i=0}^{k}{C_{k}^{i}{{x}^{i}}} \right]=\sum\limits_{k=0}^{n}{\left[ \sum\limits_{i=0}^{k}{C_{n}^{k}{{x}^{k}}C_{k}^{i}{{x}^{i}}} \right]}=\sum\limits_{k=0}^{n}{\left[ \sum\limits_{i=0}^{k}{C_{n}^{k}C_{k}^{i}{{x}^{k+i}}} \right]}.$
Số hạng của ${{x}^{3}}$ tương ứng với:
$\left\{ \begin{matrix}
k+i=3 \\
0\le k\le n \\
0\le i\le k \\
\end{matrix} \right.\Rightarrow (k,i)\in \left\{ (3;0),(2;1) \right\}$.
Do đó ${{a}_{3}}=C_{n}^{3}C_{3}^{0}+C_{n}^{2}C_{2}^{1}=C_{n}^{3}+2C_{n}^{2}$.
Số hạng của ${{x}^{4}}$ tương ứng với:
$\left\{ \begin{matrix}
k+i=4 \\
0\le k\le n \\
0\le i\le k \\
\end{matrix} \right.\Rightarrow (k,i)\in \left\{ (4;0),(3;1),(2;2) \right\}$.
Do đó ${{a}_{4}}=C_{n}^{4}C_{4}^{0}+C_{n}^{3}C_{3}^{1}+C_{n}^{2}C_{2}^{2}=C_{n}^{4}+3C_{n}^{3}+C_{n}^{2}$.
Theo bài ra: $\dfrac{{{a}_{3}}}{14}=\dfrac{{{a}_{4}}}{41}$ nên:
$\dfrac{C_{n}^{3}+2C_{n}^{2}}{14}=\dfrac{C_{n}^{4}+3C_{n}^{3}+C_{n}^{2}}{41}\Leftrightarrow 41C_{n}^{3}+82C_{n}^{2}=14C_{n}^{4}+42C_{n}^{3}+14C_{n}^{2}$
$\Leftrightarrow 14C_{n}^{4}+C_{n}^{3}-68C_{n}^{2}=0\Leftrightarrow (n-1)n\left[ \dfrac{7}{12}(n-3)(n-2)+\dfrac{n-2}{6}-34 \right]=0$
$\Leftrightarrow (n-1)n(7{{n}^{2}}-33n-370)=0\Rightarrow n=10 (do n\ge 2, n\in \mathbb{N})$
Khi đó tổng $S={{a}_{0}}+{{a}_{1}}+{{a}_{2}}+\ldots +{{a}_{2n}}=f(1)={{3}^{10}}$.
Khi đó:
$f(x)={{\left( 1+x+{{x}^{2}} \right)}^{n}}={{\left( x+{{x}^{2}}+1 \right)}^{n}}=\sum\limits_{k=0}^{n}{C_{n}^{k}}{{(x+{{x}^{2}})}^{k}}{{.1}^{n-k}}=\sum\limits_{k=0}^{n}{C_{n}^{k}}{{(x+{{x}^{2}})}^{k}}=\sum\limits_{k=0}^{n}{C_{n}^{k}}{{x}^{k}}{{(x+1)}^{k}}=$
$=\sum\limits_{k=0}^{n}{C_{n}^{k}{{x}^{k}}}\left[ \sum\limits_{i=0}^{k}{C_{k}^{i}{{x}^{i}}} \right]=\sum\limits_{k=0}^{n}{\left[ \sum\limits_{i=0}^{k}{C_{n}^{k}{{x}^{k}}C_{k}^{i}{{x}^{i}}} \right]}=\sum\limits_{k=0}^{n}{\left[ \sum\limits_{i=0}^{k}{C_{n}^{k}C_{k}^{i}{{x}^{k+i}}} \right]}.$
Số hạng của ${{x}^{3}}$ tương ứng với:
$\left\{ \begin{matrix}
k+i=3 \\
0\le k\le n \\
0\le i\le k \\
\end{matrix} \right.\Rightarrow (k,i)\in \left\{ (3;0),(2;1) \right\}$.
Do đó ${{a}_{3}}=C_{n}^{3}C_{3}^{0}+C_{n}^{2}C_{2}^{1}=C_{n}^{3}+2C_{n}^{2}$.
Số hạng của ${{x}^{4}}$ tương ứng với:
$\left\{ \begin{matrix}
k+i=4 \\
0\le k\le n \\
0\le i\le k \\
\end{matrix} \right.\Rightarrow (k,i)\in \left\{ (4;0),(3;1),(2;2) \right\}$.
Do đó ${{a}_{4}}=C_{n}^{4}C_{4}^{0}+C_{n}^{3}C_{3}^{1}+C_{n}^{2}C_{2}^{2}=C_{n}^{4}+3C_{n}^{3}+C_{n}^{2}$.
Theo bài ra: $\dfrac{{{a}_{3}}}{14}=\dfrac{{{a}_{4}}}{41}$ nên:
$\dfrac{C_{n}^{3}+2C_{n}^{2}}{14}=\dfrac{C_{n}^{4}+3C_{n}^{3}+C_{n}^{2}}{41}\Leftrightarrow 41C_{n}^{3}+82C_{n}^{2}=14C_{n}^{4}+42C_{n}^{3}+14C_{n}^{2}$
$\Leftrightarrow 14C_{n}^{4}+C_{n}^{3}-68C_{n}^{2}=0\Leftrightarrow (n-1)n\left[ \dfrac{7}{12}(n-3)(n-2)+\dfrac{n-2}{6}-34 \right]=0$
$\Leftrightarrow (n-1)n(7{{n}^{2}}-33n-370)=0\Rightarrow n=10 (do n\ge 2, n\in \mathbb{N})$
Khi đó tổng $S={{a}_{0}}+{{a}_{1}}+{{a}_{2}}+\ldots +{{a}_{2n}}=f(1)={{3}^{10}}$.
Đáp án C.