by .
Giả sử $P\left( x \right)={{\left( 2x+1 \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{n}}{{x}^{n}}$ thỏa mãn ${{a}_{0}}+\frac{{{a}_{1}}}{2}+\frac{{{a}_{2}}}{{{2}^{2}}}+…+\frac{{{a}_{n}}}{{{2}^{n}}}={{2}^{12}}$. Hệ số lớn nhất trong các hệ số $\left\{ {{a}_{0}},{{a}_{1}},{{a}_{2}},…,{{a}_{n}} \right\}$là
C. $126720$.
B. $495$.
C. $256$.
D. $591360$.
Hướng dẫn
Đáp án A
$\begin{align}
& {{2}^{12}}={{a}_{0}}+\frac{{{a}_{1}}}{2}+\frac{{{a}_{2}}}{{{2}^{2}}}+……+\frac{{{a}^{n}}}{{{2}^{n}}}={{a}_{0}}+{{a}_{1}}\left( \frac{1}{2} \right)+{{a}_{2}}{{\left( \frac{1}{2} \right)}^{2}}+…+{{a}_{n}}{{\left( \frac{1}{2} \right)}^{n}} \\
& =P\left( \frac{1}{2} \right)={{\left( 1+2.\frac{1}{2} \right)}^{n}}={{2}^{n}} \\
\end{align}$
$\Rightarrow n=12$
${{\left( 2x+1 \right)}^{12}}=\sum\limits_{k=0}^{12}{C_{12}^{k}}{{\left( 2x \right)}^{k}}=\sum\limits_{k=0}^{12}{C_{12}^{k}.{{x}^{k}}}{{2}^{k}}.$
$\Rightarrow {{a}_{k}}=C_{12}^{k}{{.2}^{k}}\forall k\overline{0,12}\Rightarrow {{a}_{k}}\le {{a}_{k+1}}\Leftrightarrow C_{12}^{k}{{.2}^{k}}\le C_{12}^{k+1}{{.2}^{k+1}}$
$\begin{align}
& \frac{12!}{k!.\left( 12-k \right)!}\le \frac{12!}{\left( k+1 \right)!.\left( 11-k \right)!} \\
& \Leftrightarrow \frac{1}{12-k}\le \frac{2}{k+1} \\
& \Leftrightarrow k\le \frac{23}{3},k\in \mathbb{N}\Rightarrow k\in \left\{ 0,1,2,3,…7 \right\} \\
\end{align}$
Từ đây ta có:
$\left\{ \begin{align}
& {{a}_{k}}\le {{a}_{k+1}}\forall k\in \left\{ 0,1,2,3,…7 \right\} \\
& {{a}_{k}}={{a}_{k+1}}\Leftrightarrow k=\frac{23}{3},k\notin N \\
& {{a}_{k}}>{{a}_{k+1}}\forall k\in \left\{ 8;9;….11 \right\} \\
\end{align} \right.$
Do đó: ${{a}_{0}}<{{a}_{1}}<{{a}_{2}}<{{a}_{3}}<{{a}_{4}}<{{a}_{5}}<…..<{{a}_{8}}>{{a}_{9}}>….>{{a}_{12}}$
Vậy ${{a}_{5}}=max\left\{ {{a}_{i}}\left| i=\overline{0,12} \right. \right\}=C_{12}^{8}{{.2}^{8}}$
