Ta có:
$(1+x)^n=\sum_{k=0}^nC^k_nx^k$
$\frac{a_{k-1}}{2}=\frac{a_k}{9}=\frac{a_{k+1}}{24}\Leftrightarrow \frac{C^{k-1}_n}{2}=\frac{C^k_n}{9}=\frac{C^{k+1}_n}{24}$
$\Leftrightarrow \frac{n!}{2(k-1)!(n-k+1)!}=\frac{n!}{9k!(n-k)!}=\frac{n!}{24(k+1)!(n-k-1)!}$
$\Leftrightarrow 2(k-1)!(n-k+1)!=9k!(n-k)!=24(k+1)!(n-k-1)!$
$\Leftrightarrow \left\{\begin{array}{l}2(n-k+1)=9k\\9(n-k)=24(k+1)\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}k=\frac{2n+2}{11}\\k=\frac{3n-8}{11}\end{array}\right.$
$\Rightarrow 2n+2=3n-8 \Leftrightarrow n=10$