I propose an easier way:

If average = sum of #s / (# of #s), it follows that average * (# of #s) = sum of #s.

Find the average by (first + last) / 2 (given that the numbers are evenly-spaced).

That is, (50+1)/2 = 51/2.

Find the number of numbers. That's 1 to 50, effectively just counting on your fingers, which is 50 numbers.

Therefore, we find average * (# of #s) = 51/2 * 50 = 51 * 25 (note here that I've actually divided the 50 by 2 rather than the 51 because the answer is the same and 25.5 is annoying to work with).

51*25 = 1275. Answer C.

So... my issue with the (n(n+1))/2 thing is that this equation assumes we're starting from 1. What does this mean?

Well, the numbers 1 to n would be n numbers. The average of the integers 1 to n would be (1+n)/2. If we put these into average * (# of #s), we find (n*(1+n)/2) = (n(n+1)/2).

Now in this case that works well because we are starting with 1. Then consider a case such as "the sum of all the even integers from 100 to 300 inclusive" and good luck with the equation, at least under GRE time constraints: you'd have to do the equation for 301 and then 99 and subtract, which is time-consuming versus just finding the average and the number of numbers.

Remember, you must understand the concept thoroughly before using any equation or the GRE will trip you (just because it can). In this case, as in many, knowing the concept actually makes the solution easier than simply plugging and chugging.