There are plenty of ways; it all depends on what kind of function you want.
One way is to use Lagrange interpolation to get a polynomial that will have the values you want (e.g., $f(1)=100$, $f(2)=250$, etc). This, however, may not give you the "correct" long-term behavior, and the long-term growth is not going to be exponential. You will "hit" all the right values for the points you are given, but you may not have good control later.
(For example, if you use Lagrange polynomial interpolation to find a function that satisfies $f(0)=1,\quad f(1)=2,\quad f(2)=4,\quad f(3)=8,\quad\text{and}\quad f(4)=16,$ then this function will give $f(5)=31$ (rather than the perhaps expected $32$).
A more likely long-term behavior will be given by exponential functions. If the rate of growth is relatively constant, then that's the way to go. Consider the sequence of quotients of successive values: $\frac{f(2)}{f(1)},\quad\frac{f(3)}{f(2)},\quad\frac{f(4)}{f(3)},\quad\frac{f(5)}{f(4)},\ldots$ and if these are all reasonably close to one another, then take $a$ to be their average. You can approximate the values pretty well with $f(n) = f(1)a^{n-1}.$ This is what Ross gives you, with $a=2.5$ and $f(1)=100$, except he's writing it as $f(n) = 100(2.5)^{n-1} = \left(\frac{100}{2.5}\right)2.5^n = 40\times 2.5^n.$
If the rate of growth is (the successive factors) are changing a lot, though, it's going to be a lot harder to come up with a good exponential approximation.