I've already said about that in the previous reply - if the solution between 0 and 1 doesn't exist, f returns unevaluated, and in all those examples that you mentioned, it is easy to see from the plots given by g, that the solution doesn't exist - the lhs is negative on that interval. For example, take a look at
plots:-display(g(10, 1/3, 2, 1),plot(0,0..1));
If you want to find a solution outside that interval, the range 0..1 should be removed from the fsolve in f, and modified this way procedure, call it f1, would give a solution outside of that interval (if it exists),
f1(10, 1/3, 2, 1);
I don't think that there is a simple formula other than with the RootOf for the symbolic solution - try to solve the expression that f(10, 1/3, 2, 1) returns inside of the unevaluated fsolve,
Maple returns it as a RootOf of a polynomial of 10th degree. The same is the general situation - the solution is a root of a polynomial of degree n, which can be explicitely written if you are interested in it, but that doesn't help with finding the root - it is generally the same equation as you had originally with fractions - it is the numerator of their sum written as one fraction.
Some asymptotics, certainly, could be found, say for n approaching infinity - perhaps even the entire series - but that's it.