Câu hỏi 11

\(A_n^2 + C_{n + 1}^{n - 1} = 4n + 6\)  \(\left( {n \ge 2;\,\,n \in N} \right)\)

\( \Leftrightarrow \dfrac{{n!}}{{\left( {n - 2} \right)!}} - \dfrac{{\left( {n + 1} \right)!}}{{2\left( {n - 1} \right)!}} = 4n + 6\)

\( \Leftrightarrow n\left( {n - 1} \right) - \dfrac{{n\left( {n + 1} \right)}}{2} = 4n + 6\)

\( \Leftrightarrow 2n\left( {n - 1} \right) - n\left( {n + 1} \right) - 8n - 12 = 0\)

\( \Leftrightarrow 2{n^2} - 2n - {n^2} - n - 8n - 12 = 0\)

\( \Leftrightarrow {n^2} - 11n - 12 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}n = 12\,\,\,\left( {tm} \right)\\n =  - 1\,\,\,\left( {ktm} \right)\end{array} \right.\)

Chọn A.