Đáp án: ${a_{13}} = C_{2018}^{13}{.2^{13}};S = {3^{2011}}$
Giải thích các bước giải:
$\begin{array}{l}
+ )P\left( x \right) = {\left( {1 + 2x} \right)^{2018}} = {a_0} + {a_1}x + ... + {a_{2018}}{x^{2018}}\\
= \sum\limits_{k = 0}^{2018} {C_{2018}^k{{.2}^k}.{x^k}} \\
\Rightarrow {a_{13}} = C_{2018}^{13}{.2^{13}}\\
+ ){\left( {1 + 2x} \right)^{2011}} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{2011}}{x^{2011}}\\
= \sum\limits_{k = 0}^{2011} {C_{2018}^k{{.2}^k}.{x^k}} \\
Tổng\,hệ\,số:S = {a_0} + {a_1} + {a_2} + ... + {a_{2011}}\\
Ứng\,với\,x = 1\\
\Rightarrow S = {\left( {1 + 2.1} \right)^{2011}} = {3^{2011}}
\end{array}$
