Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
C_n^3 + 2n = A_{n + 1}^2 \Leftrightarrow \frac{{n!}}{{3!\left( {n - 3} \right)!}} + 2n = \frac{{\left( {n + 1} \right)!}}{{\left( {n - 1} \right)!}}\,\,\,\,\left( {n \ge 3} \right)\\
\Leftrightarrow \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} + 2n = \left( {n + 1} \right).n\\
\Leftrightarrow n\left( {{n^2} - 3n + 2} \right) + 12n = 6n\left( {n + 1} \right)\\
\Leftrightarrow {n^2} - 3n + 2 + 12 = 6\left( {n + 1} \right)\\
\Leftrightarrow {n^2} - 3n + 14 = 6n + 6\\
\Leftrightarrow {n^2} - 9n + 8 = 0 \Leftrightarrow \left[ \begin{array}{l}
n = 1\left( {loai} \right)\\
n = 8
\end{array} \right.\\
{\left( {2x - \frac{3}{{\sqrt[3]{x}}}} \right)^{16}} = \sum\limits_{k = 0}^{16} {C_{16}^k{{\left( {2x} \right)}^{16 - k}}.{{\left( { - 3{x^{ - \frac{1}{3}}}} \right)}^k}} \\
= \sum\limits_{k = 0}^{16} {C_{16}^k{2^{16 - k}}.{{\left( { - 3} \right)}^k}.{x^{16 - k}}.{x^{ - \frac{k}{3}}}} = \sum\limits_{k = 0}^{16} {C_{16}^k{2^{16 - k}}.{{\left( { - 3} \right)}^k}.{x^{\frac{{48 - 4k}}{3}}}} \\
Cho\,\frac{{48 - 4k}}{3} = 0 \Leftrightarrow k = 12\,thi\,so\,hang\,khong\,chua\,x\,la:\\
C_{16}^{12}{.2^{16 - 12}}.{\left( { - 3} \right)^{12}} = {16.3^{12}}.C_{16}^{12}
\end{array}$
