Solution
= Solution
Here $a,b\geq2$ are fixed <integers>. The term $ab^n$ has <exponential growth>, whereas $an^b$ has fixed-degree <polynomial growth> and $nb^a$ is linear. Since a fixed-base exponential eventually dominates every fixed-degree polynomial,
$$
\frac{an^b+ab^n+nb^a}{b^n}\longrightarrow a.
$$
Hence the requested <Big O notation> bound, with its matching lower order, is $\boxed{\Theta(b^n),\quad\text{in particular }O(b^n)}$.