Olá, Dan.
Isto é um Binômio de Newton.
[tex](1+4)^n=\sum\limits_{k=0}^n \binom{n}{k}1^{n-k}4^{k}=\sum\limits_{k=0}^n \binom{n}{k}4^{k}=\\\\=\binom n 04^0+\binom n 14^1+\binom n 24^2+...+\binom n n4^n=625 \Rightarrow \\\\ 5^n=625 \Rightarrow 5^n=5^4\\\\ \therefore \text{Igualando os expoentes: }\boxed{n=4}[/tex]
