`~rai~`
\(\text{Câu 4:}\\A_n^2-3C_n^{n-1}=11n\\\Leftrightarrow \dfrac{n!}{(n-2)!}-3.\dfrac{n!}{(n-1)!.[n-(n-1)]!}=11n\\\Leftrightarrow \dfrac{n(n-1)(n-2)!}{(n-2)!}-3.\dfrac{n(n-1)!}{(n-1)!}=11n\\\Leftrightarrow n(n-1)-3n=11n\\\Leftrightarrow n^2-n-3n-11n=0\\\Leftrightarrow n^2-15n=0\\\Leftrightarrow n(n-15)=0\\\Leftrightarrow \left[\begin{array}{I}n=0\text{(không thỏa mãn)}\\n=15.\text{(thỏa mãn)}\end{array}\right.\\\Rightarrow (x^2+x+1)^n=(x^2+x+1)^{15}\\\text{Xét khai triển }(x^2+x+1)^{15}\\=[x^2+(x+1)]^{15}\\=\sum_{k=0}^{15}C_{15}^k(x^2)^{15-k}.(x+1)^k\\=\sum_{k=0}^{15}C_{15}^kx^{30-2k}.\sum_{l=0}^kC_k^l1^{k-l}x^l\\=\sum_{k=0}^{15}C_{15}^kx^{30-2k}.\sum_{l=0}^kC_k^lx^l\\=\sum_{k=0}^{15}C_{15}^k.\sum_{l=0}^kC_k^lx^{30-2k+l}.\\\text{Số hạng chứa }x^{12}\text{ tương ứng với (k;l) thỏa mãn:}\\30-2k+l=12\\\Leftrightarrow l=2k-18.\\\text{Kết hợp với điều kiện,ta có hệ:}\\\begin{cases}l=2k-18\\0\le k\le 15,k\in\mathbb{N}\\0\le l\le k,l\in\mathbb{N}\end{cases}\\\Leftrightarrow \begin{cases}l=2k-18\\9\le k\le 15,k\in\mathbb{N}\\0\le l\le k,l\in\mathbb{N}\end{cases}\\\Rightarrow k\in\{9;10;11;12;13;14;15\}\\\Leftrightarrow (k;l)=\{(9;0),(10;2),(11;4),(12;6),(13;8);(14,10);(15;12)\}.\text{Cộng lại.}\\\text{Vậy hệ số của }x^{11}\text{ bằng tổng:}\\C_{15}^9+C_{15}^{10}.C_{10}^2+C_{15}^{11}.C_{11}^4+C_{15}^{12}.C_{12}^6+C_{15}^{13}.C_{13}^8+C_{15}^{14}.C_{14}^{10}+C_{15}^{12}.\)
