Đáp án:
$n = 4$
Giải thích các bước giải:
\(\begin{array}{l}
\quad 2(A_n^3 + 3A_n^2) = P_{n+1}\qquad (n\geqslant 3;n\in\Bbb N)\\
\Leftrightarrow 2\cdot \dfrac{n!}{(n-3)!} + 6\cdot \dfrac{n!}{(n-2)!} = (n+1)!\\
\Leftrightarrow \dfrac{2}{(n-3)!} + \dfrac{6}{(n-2)!} = n+1\\
\Leftrightarrow \dfrac{2}{(n-3)!}\left(1 + \dfrac{3}{n-2}\right) = n + 1\\
\Leftrightarrow \dfrac{2}{(n-3)!}\cdot \dfrac{n+1}{n-2} = n+1\\
\Leftrightarrow \dfrac{2}{(n-2)!} = 1\qquad (Do\ n+1 \ne 0)\\
\Leftrightarrow (n-2)! = 2\\
\Leftrightarrow (n-2)! = 2!\\
\Leftrightarrow n - 2= 2\\
\Leftrightarrow n = 4\quad \text{(nhận)}\\
\text{Vậy}\ n=4
\end{array}\)
