You can certainly ignore the information given in the question about the sum of the first 50 positive evens, and apply a formula directly to get an answer. It's quite quick as well, so not a bad approach. Still, you can get the answer without the formula, by using the information given:

2+4+6+...+98+100 = 2550

102+104+106+...+198+200 =
100+2+100+4+100+6+...+100+98+100+100 =
2+4+6+...+98+100+50*100 =
2550 + 5000 =
7550

VerbalAttack wrote: Comments most welcome..

Yes, your method is perfectly correct. There are several ways to solve these questions, and as long as you have a method you can apply that gets an answer within a minute or a minute and a half, you should be fine. I'll explain how I approach these questions below, but I don't want to suggest that the method below is any better than the method above- it's just different, and doesn't need a formula, but it does require a bit of thought about how to rearrange each sum.

As long as you know how to add

1+2+3+...+49+50

you can add any sum of numbers in an arithmetic progression (i.e. a sequence that increases by the same amount between each term). I'll call this kind of sum an 'evenly spaced' sum, since consecutive numbers are always the same distance apart. The key is to add the smallest and largest (1+50 = 51), and notice that you get the same thing when you add the second largest and the second smallest (2+49=51), and so on.

So
1+2+3+...+49+50 = 51+51+51...+51

And how many 51s are we adding? Well, we need to know how many pairs of numbers there are in the list 1, 2, 3, ..., 49, 50. There are clearly 50 numbers, and therefore 50/2 = 25 pairs. So

1+2+3+...+49+50 = 51+51+51...+51 = 25*51 = 1275.

I'd guess the above is already well known to most GMAT test takers. We can, however, rewrite any evenly spaced sum so that it starts at 1 and goes up by 1 each time- just like the above. To take the example from VerbalAttack's post:

What is
112+119+126+...+189+196 = ?

We can factor out a 7:

= 7*(16+17+18+...+27+28)

Now, rewrite each term in brackets as "15+x" so the sequence can start from 1, not from 16:

=7*[(15+1) + (15+2) + (15+3) + ... + (15+12)+(15+13)]
=7*(15n + 1+2+3+...+12+13)

where n is the number of terms in the original sequence. Fortunately, at this stage it is clear that n is 13. Now, using the same principle as was used above, when adding 1+2+...+49+50:

=7*(15*13 + 14*(13/2))
=7*286
=2002

I prefer not to rely on formulas, for many reasons, so the above method suits me best, and it's very quick once you practice it. But for the GMAT there are many ways to solve these types of questions, and test-takers should choose the method they're most comfortable with.