Tìm giá trị

- Hướng dẫn giải

Đáp án là A

Điều kiện: $n\ge 2$ và $n\in N$

Ta có 

$$\begin{gathered} C_{n+1}^1+3C_{n+2}^2=C_{n+1}^3 \\ \iff \frac{(n+1)!}{1!.n!}+3.\frac{(n+2)!}{2!.n!}=\frac{(n+1)!}{3!.(n-2)!} \end{gathered}$$

$$\begin{gathered} \iff n+1+3.\frac{(n+1).(n+2)}{2}=\frac{(n-1).n.(n+1)}{6} \\ \iff 1+3.\frac{(n+2)}{2}=\frac{(n-1).n}{6} \end{gathered}$$ 

( vì n + 1 > 0 nên chia cả 2 vế cho n +1).

$$\begin{gathered} \iff 6 +\hspace{0.17em}3.3(n +\hspace{0.17em}2)= (n-1).n \\ \iff 6+9n+18=n^2-n \\ \iff n^2-10n-24=0 \\ \iff \left\{\begin{array}{l}n=-2\left(L\right)\\ n=12\left(tm\right)\end{array}\right. \end{gathered}$$