Chọn phương án A.

Điều kiện: \(n\geq2\).
$$\begin{eqnarray*}
&\mathrm{C}_{n+1}^1+3\mathrm{C}_{n+2}^2&=\mathrm{C}_{n+1}^3\\
\Leftrightarrow&\dfrac{(n+1)!}{1!\cdot n!}+\dfrac{3(n+2)!}{2!\cdot n!}&=\dfrac{(n+1)!}{3!(n-2)!}\\
\Leftrightarrow&\dfrac{(n+1)n!}{n!}+\dfrac{3(n+2)(n+1)n!}{2n!}&=\dfrac{(n+1)n(n-1)(n-2)!}{6(n-2)!}\\
\Leftrightarrow&(n+1)+\dfrac{3(n+2)(n+1)}{2}&=\dfrac{(n+1)n(n-1)}{6}\\
\Leftrightarrow&6(n+1)+9(n+2)(n+1)&=(n+1)n(n-1)\\
\Leftrightarrow&6+9(n+2)&=n(n-1)\\
\Leftrightarrow&n^2-10n-24&=0\\
\Leftrightarrow&\left[\begin{array}{ll}x=-2 &\text{(loại)}\\ x=12 &\text{(nhận)}\end{array}\right.
\end{eqnarray*}$$