Consider the typical problem,
You have a conducting sphere of charge $Q$ and a point charge $q$ a certain distance away, what is the force on the point charge?
The solution is a simple application of the method of images, but in Griffiths part of the argument relies on saying
$$Q_\text{conductor} = \sum_i q_{\ \text{image}}$$
Or, that the image charges in the conductor add to the total charge. I don't really see how this is a mathematically valid statement. I was thinking it may come from the Neumann boundary conditions and integrating Poisson's eq, but I am not sure.
For reference consider Griffiths Problem 3.8 which deals with the $Q=0$ case.
An explanation would really be helpful.
Best Answer
The formula for $Q_{conductor}$ that you reference comes from the mid-way point of an argument from superposition:
1) Start by assuming the conducting sphere is grounded. (i.e. forget about the charge $Q$ for now.)
a) Use the standard method of images to replace the grounded sphere with an equivalent image charge $q_{image} = -q (a/r)$ at position $a^2/r$, where a is the radius of the conducting sphere and $r$ is the distance of charge $q$ from the center of the sphere. You can now find the field.
b) Return to the grounded sphere. The (now known) field requires charge on the surface of the sphere to terminate it. By Gauss' law, the total charge on the surface, $Q_{conductor}=q_{image}$. That's your formula (which is generalized to the case of multiple image charges).
c) You can now break the connection grounding the sphere. Nothing changes; the sphere surface remains an equipotential at ground level.
2) With the sphere no longer grounded, you can now add charge $Q-q_{image}$ to the sphere to achieve the original problem specification. Since the electrostatic forces were balanced in part 1), the new charge distributes itself uniformly over the sphere, and so acts like a point charge at the center of the sphere.
You can now add the two fields from parts 1) and 2) to find the force on charge $q$.