Tìm n biết \(A_n^3 + C_n^{n - 2} = 14n\) với n > 2, n ∈ ℕ.

Hướng dẫn giải

Đáp án đúng là: B

Ta có:  \(A_n^3 + C_n^{n - 2} = 14n\)

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

\( \Leftrightarrow \frac{{n.(n - 1).(n - 2).(n - 3)!}}{{(n - 3)!}} + \frac{{n.(n - 1)(n - 2)!}}{{(n - 2)!.2!}} = 14n\)

\( \Leftrightarrow n(n - 1)(n - 2) + \frac{{n(n - 1)}}{2} = 14n\)

\( \Leftrightarrow ({n^2} - n).(n - 2) + \frac{{{n^2} - n}}{2} = 14n\)

\( \Leftrightarrow {n^3} - {n^2} - 2{n^2} + 2n + \frac{{{n^2} - n}}{2} = 14n\)

\( \Leftrightarrow 2{n^3} - 2{n^2} - 4{n^2} + 4n + {n^2} - n - 28n = 0\)

\( \Leftrightarrow 2{n^3} - 5{n^2} - 25n = 0\)

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

\( \Leftrightarrow n(2n + 5)(n - 5) = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}n = 0\\2n + 5 = 0\\n - 5 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}n = 0(ktm)\\n = \frac{{ - 5}}{2}(ktm)\\n = 5(tm)\end{array} \right.\)

Vậy n = 5.