Cho khai triển ${{\left( 1+2x...

Ta viết ${{\left( 1+2x \right)}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}=\sum\limits_{k=0}^{n}{{{a}_{k}}{{x}^{k}}}$. Lại có: ${{\left( 1+2x \right)}^{n}}=\sum\limits_{k=0}^{n}{C_{n}^{k}{{2}^{k}}{{x}^{k}}}$ nên ${{a}_{k}}=C_{n}^{k}{{2}^{k}}$. Vì vậy ${{a}_{0}}+\dfrac{{{a}_{1}}}{2}+...+\dfrac{{{a}_{n}}}{{{2}^{n}}}=4096$ hay
$\sum\limits_{k=o}^{n}{\dfrac{{{a}_{k}}}{{{2}^{k}}}}=4096\Leftrightarrow \sum\limits_{k=o}^{n}{\dfrac{C_{n}^{k}{{2}^{k}}}{{{2}^{k}}}}=4096\Leftrightarrow \sum\limits_{k=o}^{n}{C_{n}^{k}}=4096\Leftrightarrow {{\left( 1+1 \right)}^{n}}=4096\Leftrightarrow {{2}^{n}}=4096\Leftrightarrow n=12$.
Suy ra ${{a}_{k}}=C_{12}^{k}{{2}^{k}}$, $k=\overline{0,12}$. Nếu ${{a}_{k}}$ lớn nhất thì:
$\left\{ \begin{aligned}
& {{a}_{k}}\ge {{a}_{k+1}} \\
& {{a}_{k}}\ge {{a}_{k-1}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& C_{12}^{k}{{2}^{k}}\ge C_{12}^{k+1}{{2}^{k+1}} \\
& C_{12}^{k}{{2}^{k}}\ge C_{12}^{k-1}{{2}^{k-1}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{12!}{k!\left( 12-k \right)!}{{2}^{k}}\ge \dfrac{12!}{\left( k+1 \right)!\left( 11-k \right)!}{{2}^{k+1}} \\
& \dfrac{12!}{k!\left( 12-k \right)!}{{2}^{k}}\ge \dfrac{12!}{\left( k-1 \right)!\left( 13-k \right)!}{{2}^{k-1}} \\
\end{aligned} \right.$;
$\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{12-k}\ge \dfrac{2}{k+1} \\
& \dfrac{2}{k}\ge \dfrac{1}{13-k} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& k\ge \dfrac{23}{3} \\
& k\le \dfrac{25}{3} \\
\end{aligned} \right.\xrightarrow{k=\overline{0,12}}k=8 $. Vậy hệ số lớn nhất là $ {{a}_{8}}=C_{12}^{8}{{.2}^{8}}=126720$.