Đáp án:
$n = 13$
Giải thích các bước giải:
$\begin{array}{l}C_{n-1}^4 = C_{n-1}^3 + \dfrac{5}{2}A_{n-2}^2\qquad (n \geq 5;\, n \in \Bbb N)\\ \Leftrightarrow C_{n-1}^3+ C_{n-1}^4 = 2C_{n-1}^3 + \dfrac{5}{2}A_{n-2}^2\\ \Leftrightarrow C_n^4 = 2C_{n-1}^3 + \dfrac{5}{2}A_{n-2}^2\\ \Leftrightarrow \dfrac{n!}{4!(n-4)!} = \dfrac{2(n-1)!}{3!(n-4)!} + \dfrac{5}{2}\cdot\dfrac{(n-2)!}{(n-4)!}\\ \Leftrightarrow \dfrac{n(n-1)(n-2)!}{4!} = \dfrac{2(n-1)(n-2)!}{3!} + \dfrac{5}{2}\cdot(n-2)!\\ \Leftrightarrow n(n-1) = 8(n-1) + 60\\ \Leftrightarrow n^2 - 9n - 52 = 0\\ \Leftrightarrow \left[\begin{array}{l}n = -4\quad (loại)\\n = 13\quad (nhận)\end{array}\right.\\ Vậy\,\,n=13 \end{array}$
