Giải bài 2.25 trang 38

Đề bài

Khai triển đa thức ${\left(1+2x\right)}^{12}$ thành dạng $a_0+a_{1}x+a_2{x}^2+...+a_{12}{x}^{12}$.

Tìm hệ số $a_k$ lớn nhất?

Lời giải chi tiết

Ta có:

$\begin{array}{c}{\left(1+2x\right)}^{12}=C_{12}^0+C_{12}^{1}2x+C_{12}^2{\left(2x\right)}^2+...+C_{12}^{12}(2x{)}^{12}\\ \Rightarrow a_k=2^k{C}_{12}^k\end{array}$

Để $a_k$ lớn nhất thì $a_{k-1}\le a_k\ge a_{k+1}\forall k$

$\begin{array}{c}\iff 2^{k-1}{C}_{12}^{k-1}\le 2^k{C}_{12}^k\ge 2^{k+1}{C}_{12}^{k+1}\\ \iff \frac{12!}{(k-1)!\left(13-k\right)!}\le 2\frac{12!}{k!\left(12-k\right)!}\ge 2^2\frac{12!}{(k+1)!\left(11-k\right)!}\\ \iff \frac{1}{\left(13-k\right)(12-k)}\le 2.\frac{1}{k\left(12-k\right)}\ge 2^2\frac{1}{k(k+1)}\\ \iff \{\begin{array}{c}\frac{1}{\left(13-k\right)(12-k)}\le 2.\frac{1}{k\left(12-k\right)}\\ 2.\frac{1}{k\left(12-k\right)}\ge 2^2\frac{1}{k(k+1)}\end{array}\\ \iff \{\begin{array}{c}\frac{1}{13-k}\le \frac{2}{k}\\ \frac{1}{12-k}\ge \frac{2}{k+1}\end{array}\iff \{\begin{array}{c}k\le 2.(13-k)\\ k+1\ge 2.(12-k)\end{array}\\ \iff \frac{23}{3}\le k\le \frac{26}{3}\Rightarrow k=8(k\in N)\end{array}$

Vậy $a_8$ là lớn nhất.