Đáp án:
$\begin{array}{l}
C_{2n + 1}^0 + C_{2n + 1}^1 + C_{2n + 1}^2 + ... + C_{2n + 1}^n = \frac{{{2^{2n + 1}}}}{2} = {2^{2n}}\\
\Rightarrow C_{2n + 1}^1 + C_{2n + 1}^2 + ... + C_{2n + 1}^n = {2^{2n}} - C_{2n + 1}^0 = {2^{2n}} - 1\\
\Rightarrow {2^{2n}} - 1 = {2^{20}} - 1\\
\Rightarrow n = 10\\
\Rightarrow {\left( {\frac{1}{{{x^4}}} + {x^7}} \right)^{10}} = \sum\limits_{k = 0}^{10} {C_{10}^k.{x^{ - 4\left( {10 - k} \right)}}.{x^{7k}} = \sum\limits_{k = 0}^{10} {C_{10}^k.{x^{11k - 40}}} } \\
{x^{26}} \Rightarrow 11k - 40 = 26 \Rightarrow 11k = 66 \Rightarrow k = 6\\
\Rightarrow Hệ\,số:C_{10}^6 = 210\\
b){\left( {1 + 3x + 2{x^2}} \right)^{10}}\\
= \sum\limits_{k = 0}^{10} {C_{10}^k.{{\left( {2{x^2}} \right)}^{10 - k}}.{{\left( {1 + 3x} \right)}^k}} \\
= \sum\limits_{k = 0}^{10} {.\sum\limits_{m = 0}^k {C_{10}^k.C_k^m{{.2}^{10 - k}}{{.3}^m}.{x^{20 - 2k + m}}} } \left( {0 \le m \le k \le 10} \right)\\
{a_{15}} \Rightarrow {x^{15}} \Rightarrow 20 - 2k + m = 15\\
\Rightarrow 2k - m = 5\\
\Rightarrow \left[ \begin{array}{l}
k = 3;m = 1\\
k = 4;m = 3\\
k = 5;m = 5
\end{array} \right.\\
\Rightarrow {a_{15}} = C_{10}^k.C_k^m{.2^{10 - k}}{.3^m} = 3549312
\end{array}$
