Lời giải chi tiết:

\(A_n^3 + C_n^{n - 2} = 14n\)\(\left\{ \begin{array}{l}n \ge 3\\n \in N\end{array} \right.\)

\( \Leftrightarrow \dfrac{{n!}}{{\left( {n - 3} \right)!}} + \dfrac{{n!}}{{\left( {n - 2} \right)!.2!}} = 14n\)

\( \Leftrightarrow \left( {n - 2} \right)\left( {n - 1} \right)n + \dfrac{{\left( {n - 1} \right)n}}{{2!}} = 14n\)

\( \Leftrightarrow \left( {n - 2} \right)\left( {n - 1} \right) + \dfrac{{n - 1}}{2} = 14\) (Chia 2 vế cho n vì \(n \ne 0\))

\( \Leftrightarrow 2{n^2} - 5n - 25 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}n = 5\,\,\,\,\,\,\,\,\,\,\left( {tm} \right)\\n =  - 2,5\,\,\,\left( {ktm} \right)\end{array} \right.\)

Chọn A.