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:
\(\begin{array}{l}{\left( {1 + 2x} \right)^{12}} = C_{12}^0 + C_{12}^12x + C_{12}^2{\left( {2x} \right)^2} + ... + C_{12}^{12}{(2x)^{12}}\\ \Rightarrow {a_k} = {2^k}C_{12}^k\end{array}\)
Để \({a_k}\) lớn nhất thì \({a_{k - 1}} \le {a_k} \ge {a_{k + 1}}\forall k\)
\(\begin{array}{l} \Leftrightarrow {2^{k - 1}}C_{12}^{k - 1} \le {2^k}C_{12}^k \ge {2^{k + 1}}C_{12}^{k + 1}\\ \Leftrightarrow \frac{{12!}}{{(k - 1)!\left( {13 - k} \right)!}} \le 2\frac{{12!}}{{k!\left( {12 - k} \right)!}} \ge {2^2}\frac{{12!}}{{(k + 1)!\left( {11 - k} \right)!}}\\ \Leftrightarrow \frac{1}{{\left( {13 - k} \right)(12 - k)}} \le 2.\frac{1}{{k\left( {12 - k} \right)}} \ge {2^2}\frac{1}{{k(k + 1)}}\\ \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{{\left( {13 - k} \right)(12 - k)}} \le 2.\frac{1}{{k\left( {12 - k} \right)}}\\2.\frac{1}{{k\left( {12 - k} \right)}} \ge {2^2}\frac{1}{{k(k + 1)}}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{{13 - k}} \le \frac{2}{k}\\\frac{1}{{12 - k}} \ge \frac{2}{{k + 1}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}k \le 2.(13 - k)\\k + 1 \ge 2.(12 - k)\end{array} \right.\\ \Leftrightarrow \frac{{23}}{3} \le k \le \frac{{26}}{3} \Rightarrow k = 8\;(k \in \mathbb{N})\end{array}\)
Vậy \({a_8}\) là lớn nhất.
English
French
Spanish
Russian