Cho khai triển nhị thức (1/3 +2/3 x)^10 = a0 +a1 x +a2 x^2

* Hướng dẫn giải

Đáp án D

Số hạng tổng quát của khai triển: $C_{10}^k{\left(\frac{1}{3}\right)}^{10-k}.{\left(\frac{2}{3}x\right)}^k=C_{10}^k\frac{2^k}{3^{10}}{x}^k=a_k.x^k\left(k\ge 0\right)$.

Giả sử a_k là hệ số lớn nhất thì 

$\begin{array}{l}\left\{\begin{array}{l}{a}_k\ge a_{k+1}\\ a_k\ge a_{k-1}\end{array}\right.\iff \left\{\begin{array}{l}{C}_{10}^k\frac{2^k}{3^{10}}\ge C_{10}^{k+1}\frac{2^{k+1}}{3^{10}}\\ C_{10}^k\frac{2^k}{3^{10}}\ge C_{10}^{k-1}\frac{2^{k-1}}{3^{10}}\end{array}\right.\\ \iff \left\{\begin{array}{l}\frac{10!}{\left(10-k\right)!.k!}\ge \frac{2.10!}{\left(9-k\right)!.\left(k+1\right)!}\\ \frac{2.10!}{\left(10-k\right)!.k!}\ge \frac{10!}{\left(11-k\right)!.\left(k-1\right)!}\end{array}\right.\\ \iff \left\{\begin{array}{l}\frac{1}{10-k}\ge \frac{2}{k+1}\\ \frac{2}{k}\ge \frac{1}{11-k}\end{array}\right.\iff \left\{\begin{array}{l}k+1\ge 2\left(10-k\right)\\ 2\left(11-k\right)\ge k\end{array}\right.\\ \iff \left\{\begin{array}{l}k\ge \frac{19}{3}\\ k\le \frac{22}{3}\end{array}\right.\Rightarrow k=7\end{array}$