This is a B I G question.
First, let's move away from the "how much brighter..." because the answer you are looking for isn't just about brightness (which is the signal). The real answer lies in your signal to noise ratio (S/N).
Here's the concept in short form:
* There are two primary sources of noise for you to deal with. One is readout noise: the uncertainty in the values read out from the CCD sensor. The other is the shot noise - the uncertainty in the photon values arriving at the CCD sensor.
* Read noise is the uncertainty in the electron count as each pixel is read. Let's say, for the sake of simplicity, that the read noise is 10 electrons (10e- for short). This means that if you read 11,205 electrons out of a pixel, the actual value that was in there will be somewhere in the range from 11,200 to 11,210e-. So if you had four pixels, all with 11,205e-, they will probably read out with different values. These values might be 11,209; 11,201; 11,205, and 11,210. This difference in values creates the graininess that you see in an image.
* Shot noise is inherent in the physics of light. The uncertainty in the arriving photon flux is equal to the square root of the incoming signal. So if you have 400 incoming photons, the uncertainty is sqrt(400) = 20. In simplest terms, if you determine that there 400 photons have been recorded in your pixel, then you only know that the true value lies somewhere between 390 and 410 photons.
* Noise combines. So if you have some read noise and some shot noise, these two noise sources will combine to give you the total noise in the image. Noise is not simply additive, however, which means the calculation is a little complex if you don't have a math background. To determine the total noise, we use a root mean square (RMS) calculation: sum the squares of the noise sources, then take the square root of that sum. For example, we know that the shot noise is 20 and the read noise is 10. To determine the total noise:
sqrt(20^2 + 10^) = sqrt(400 + 100) = sqrt(500) = 22.4 photons
(For the sake of simplicity, I've assumed that the 10e- noise of the camera is one-for-one photons for electrons - that is, the gain is 1.0.)
* Since read noise is constant, and shot noise grows with the signal level, eventually shot noise will dominate the read noise. The longer your exposure, the more the shot noise dominates the read noise. It is generally accepted that if you can get the read noise down to 5% or less of the total noise, such an exposure is said to be shot noise limited: the shot noise limits the S/N because the read noise it essentially too small to be taken into account.
* There are two sources of shot noise: the wanted signal from the object you are imaging, and the unwanted signal from sky background brightness. The brighter your sky, the greater the unwanted shot noise is. Unwanted shot noise, however, is just like the desired shot noise: it swamps the read noise, too.
* Once your shot noise in individual sub-exposures is large enough to swamp the read noise (that is, make the read noise such a small component of the total noise that it is not significant), you can add sub-exposures and get very nearly the same result you would get with a single very long exposure.
Given all of this information, it follows that you can come up with some optimal sub-exposure time for your particular combination of sky brightness and camera readout noise such that even the background of your images is shot-noise limited. There is a calculator for just this on the CCDWare web site under resources, which you can use to determine the optimal sub-exposure time for your particular conditions.
That's the Big Answer to the question implied by your request. If you want more detail, I can suggest getting a copy of my Zone System book. Anacortes stocks it.
Now, as for sum and average. There is no difference! (This illustrates how non-intuitive noise really is...)
What is an average? It's simply a scaled sum! The only potential problem you run into is if the precision in your sum is such that the scaling reduces the precision. Most CCD cameras only have 14-15 bits of dynamic range, but you are using no smaller than 16-bit numbers to contain the scaled result. Only if you have enough images to generate more than 16 bits of information do you need to worry. And even then the solution is extremely simple: just switch to floating point or 32-bit integers to save your scaled result - then there is no difference whatsoever between a sum and an average.
As it happens, as long as you retain the _real_ precision of your data (which is smaller than the 16-bit container your software typically uses), an average is the gold standard for lowest noise. (except for outliers, which gets us into a WHOLE other area of discussion and now my fingers are tired.)
Looking at the histograms, it appears to be slight but noticeable. You can test my theory by increasing the green chanel in Photoshop.
