Cho (x + 2)ⁿ = a₀ + a₁x + a₂x² +...+ a_nxⁿ. Tìm a_n để a₅ : a₆ = 12 : 7.

Ta có:  ${(x+2)}^n=\sum _{k=0}^n{C}_n^k{x}^{n-k}{2}^k$ 

 $\Rightarrow \left\{\begin{array}{l}{a}_5=C_n^5{2}^5\\ a_6=C_n^6{2}^6\end{array}\right.$

Khi đó, ta có:  $\frac{a_5}{a_6}=\frac{12}{7}\iff \frac{C_n^5{2}^5}{C_n^6{2}^6}=\frac{12}{7}$

$\iff \frac{\frac{n!}{5!(n-5)!}}{\frac{n!}{6!(n-6)!2}}=\frac{12}{7}\iff \frac{6!(n-6)!2}{5!(n-5)!}=\frac{12}{7}$

 $\iff \frac{6.2}{n-5}=\frac{12}{7}\iff n=12$

 $a_n=C_n^n{2}^n=C_{12}^{12}{2}^{12}=4096$

Vậy a_n = 4 096.