Đáp án: $240$
Giải thích các bước giải:
Ta có: $a_0+a_1+a_2+a_3+....+a_n=729$
$\to (1+2\cdot 1)^n=729$
$\to (1+2\cdot 1)^n=729$
$\to 3^n=729$
$\to n=6$
$\to a_i=C^i_6\cdot (1)^{6-i}\cdot 2^i$
$\to a_i=C^i_6\cdot 2^i$
Để $a_i$ lớn nhất
$\to \begin{cases} a_i\ge a_{i+1}\\ a_i\ge a_{i-1}\end{cases}$
$\to \begin{cases} C^i_6\cdot 2^i\ge C^{i+1}_6\cdot 2^{i+1}\\ C^i_6\cdot 2^i\ge C^{i-1}_6\cdot 2^{i-1}\end{cases}$
$\to \begin{cases} C^i_6\ge 2C^{i+1}_6\\ C^i_6\cdot 2\ge C^{i-1}_6\end{cases}$
$\to \begin{cases} \dfrac{6!}{i!\cdot (6-i)!}\ge 2\cdot \dfrac{6!}{(i+1)!\cdot (6-(i+1))!}\\ \dfrac{6!}{i!\cdot (6-i)!}\cdot 2\ge \dfrac{6!}{(i-1)!\cdot (6-(i-1))!}\end{cases}$
$\to \begin{cases} \dfrac{6!}{i!\cdot (6-i)!}\ge 2\cdot \dfrac{6!}{(i+1)!\cdot (5-i)!}\\ \dfrac{6!}{i!\cdot (6-i)!}\cdot 2\ge \dfrac{6!}{(i-1)!\cdot (7-i)!}\end{cases}$
$\to \begin{cases} \dfrac{1}{6-i}\ge 2\cdot \dfrac{1}{i+1}\\ \dfrac{1}{i}\cdot 2\ge \dfrac{1}{7-i}\end{cases}$
$\to \begin{cases}4\le i<6\\ 0<i\le 4\end{cases}$
$\to i=4$
$\to$Hệ số max là: $a_4=C^4_6\cdot 1^{6-4}\cdot 2^4=240$
