Big O constant specification Let's consider classic big O notation definition (proof link):
The $O(f(n))$ is the set of all functions such that there exist positive constants $C$ and $n_0$ with 
$$|g(n)| ≤ C * f(n)$$ 
for all $n ≥ n_0$

According to this definition it is legal to do the following:
$$g(n) = 9999 * N^2 + N = O(9999*N^2)$$
And it is also legal to note function as:
$$g(n) = 9999 * N^2 + N = O(N^2)$$
As you can see the first variant $O(9999*N^2)$ is much more precise than the second one.
The question is: why nobody use the first variant? 
 A: The definition of $g(n) \in O(f(n))$ tells us this is a class (set) of functions.  So $O(n^2) = O(9999n^2)$ because these are equal as sets (any function belonging to the left side belongs to the right side, and conversely).
To give a more "precise" (narrower) description requires a different sort of notation.  Where an author finds it important or useful to identify the "leading term" of a function (the term whose magnitude dominates as $n\to \infty$), that can be expressed by writing a sum of the leading term plus a "residual" component.
For example, your function $g(n) = 9999n^2 + n$ belongs to $O(n^2)$, but it also belongs to a strictly smaller set of functions:
$$ g(n) = 9999n^2 + n \in 9999n^2 + O(n) $$
Here the set on the right hand side is defined by adding the specific function $9999n^2$ to each element of the function class $O(n)$.  Of course one of these results is your $g(n)$.
A less precise result (but still narrower than just $O(n^2)$) would be to assert:
$$ g(n) = 9999n^2 + o(n^2) $$
This invokes the function class $o(n^2)$, defined more generally by:
$$ g(n) \in o(f(n))\; \text{ if and only if } \;\lim_{n\to \infty} \frac{g(n)}{f(n)} = 0 $$
Intuitively $o(n^2)$ means those functions which become arbitrarily small in ratio to $n^2$ (as $n$ grows arbitrarily large).
To summarize, your function $g(n)$ can be described more precisely by pulling out the leading term $9999n^2$ and estimating the growth of what remains (the residual term), namely:
$$ g(n) - 9999n^2 = n \in O(n) \subset o(n^2) $$
