. Cho số nguyên $n\ge 3$. Giả sử ta có khai triển ${{\left( x-1 \right)}^{2n}}+x{{\left( x+1 \right)}^{2n-1}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^

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Cho số nguyên $n\ge 3$. Giả sử ta có khai triển ${{\left( x-1 \right)}^{2n}}+x{{\left( x+1 \right)}^{2n-1}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{2n}}{{x}^{2n}}$. Biết$T={{a}_{0}}+{{a}_{2}}+…+{{a}_{2n}}=768.$Tính ${{a}_{5}}$.
C. $126{{x}^{5}}$.
B. $-126{{x}^{5}}$.
C. $126$.
D. $-126$.

Đáp án D
Theo giả thiết ta có:
$P\left( x \right)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{2n}}{{x}^{2n}}$.
Khi đó $P\left( 1 \right)={{a}_{0}}+{{a}_{1}}+{{a}_{2}}+…+{{a}_{2n}}$và $P\left( -1 \right)={{a}_{0}}-{{a}_{1}}+{{a}_{2}}-…+{{a}_{2n}}$.
Suy ra $T={{a}_{0}}+{{a}_{2}}+…+{{a}_{2n}}=\dfrac{P\left( 1 \right)+P\left( -1 \right)}{2}=\dfrac{{{2}^{2n-1}}+{{2}^{2n}}}{2}={{3.2}^{2n-2}}$
$\Rightarrow 768={{3.2}^{2n-2}}\Leftrightarrow n=5$
Theo công thức khai triển nhị thức Newton ta có:
$P\left( x \right)={{\left( x-1 \right)}^{2n}}+x{{\left( x+1 \right)}^{2n-1}}=\sum\limits_{k=1}^{2n}{C_{2n}^{k}{{x}^{k}}{{\left( -1 \right)}^{2n-k}}+x}\sum\limits_{k=1}^{2n-1}{C_{2n-k}^{k}{{x}^{k}}}$
$=\sum\limits_{k=1}^{2n}{C_{2n}^{k}{{x}^{k}}{{\left( -1 \right)}^{2n-k}}+}\sum\limits_{k=1}^{2n}{C_{2n-k}^{k-1}{{x}^{k}}}=1+\sum\limits_{k=1}^{2n}{\left( C_{2n}^{k}{{\left( -1 \right)}^{k}}+C_{2n-1}^{k-1} \right)}{{x}^{k}}=1+\sum\limits_{k=1}^{10}{\left( C_{10}^{k}{{\left( -1 \right)}^{k}}+C_{9}^{k-1} \right)}{{x}^{k}}$.
Vậy ${{a}_{5}}=C_{10}^{5}{{\left( -1 \right)}^{5}}+C_{9}^{4}=-126.$