 Solution

Here $a,b\ge 2$ are fixed integers. The term $a{b}^n$ has exponential growth, whereas $a{n}^b$ has fixed-degree polynomial growth and $n{b}^a$ is linear. Since a fixed-base exponential eventually dominates every fixed-degree polynomial,Hence the requested Big O notation bound, with its matching lower order, is $\boxed{\Theta (b^n),\text{in particular }O(b^n)}$.