$(5+\sqrt{x})^n=\sum_{k=1}^{n}C^{k}_{n}.5^{n-k}.x^{\frac{k}{2}} $Có $a_n$ là hệ số của $x$, vì vậy $\frac{k}{2}=1\Rightarrow k=2\Rightarrow a_n=C^{2}_{n}.5^{n-2}$
$A=\sum_{j=2}^{n}\frac{5^j}{a_j}=\sum_{j=2}^{n}\frac{5^j}{C^{2}_{j}.5^{j-2}}=\sum_{j=2}^{n}\frac{5^2}{C^{2}_{j}}$
Tính $C^{2}_{j}=\frac{j!}{(j-2)!.2!}=\frac{j.(j-1)}{2}$
Vậy $A=\sum_{j=2}^{n}\frac{5^2.2}{j.(j-1)}=50.\sum_{j=2}^{n}\frac{1}{j.(j-1)}=50.B$
Tính $B=\sum_{j=2}^{n}\frac{1}{j.(j-1)}=\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{(n-1).(n-2)}+\frac{1}{(n-1).n} $
$B=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n-2}-\frac{1}{n-1})+(\frac{1}{n-1}-\frac{1}{n})$
$B=1-\frac{1}{n}$
Vậy $A=50.(1-\frac{1}{n})=48\Rightarrow n=25$