Đáp án:
$1$ n
Giải thích các bước giải:
$C_n^6 +3C_n^7 + 3C_n^8 + C_n^9 = 2C_{n+2}^8\qquad (n \geq 9;\, n\in\Bbb N)$
$\to (C_n^6 + C_n^7)+ 2(C_n^7 + C_n^8) + (C_n^8 + C_n^9) = 2C_{n+2}^8$
$\to C_{n+1}^7 + 2C_{n+1}^8 + C_{n+1}^9 = 2C_{n+2}^8$
$\to (C_{n+1}^7 + C_{n+1}^8) + (C_{n+1}^8 + C_{n+1}^9) = 2C_{n+2}^8$
$\to C_{n+2}^8 + C_{n+2}^9 = 2C_{n+2}^8$
$\to C_{n+3}^9 = 2C_{n+2}^8$
$\to \dfrac{(n+3)!}{9!(n-6)!}=\dfrac{2(n+2)!}{8!(n-6)!}$
$\to \dfrac{(n+3)!}{(n+2)!}=\dfrac{2.9!}{8!}$
$\to n+3 = 18$
$\to n = 15$
