szy wrote:I don't know what I'm doing wrong!!

Ok so the formula I have for finding the sum of consecutive numbers is
[n(n+1)]/2

I tested this formula out by using an easy range of numbers: 2 to 8

so the total number of numbers from 2 through 8 inclusive is: (8-2) + 1 = 7

So now I plug it into the formula: [(7(7+1)]/2 --> 56/2 = 28

so the sum of the integers from 2 to 8 inclusive is 28.....

HOWEVER

I checked this by adding 2+3+4+5+6+7+8 and that gives me 35.

What am I doing wrong????

The formula you mention allows you to calculate the sum 1+ 2 + 3 + ... + n. That is, it gives the sum of the integers from 1 up to n, inclusive. If you don't start at 1, you'd need to modify the calculation somehow. There are quite a few ways this can be done. For example, if you want to find the sum 2 + 3 + 4 + ... + 8, well, that's almost the same thing as the sum 1+2+3+...+8, except that we don't want to include the 1. So if you use your formula with n=8, then subtract 1, you'll get the right answer: [8(8+1)/2] - 1 = 36 - 1 = 35.

There's a more general method to sum any 'equally spaced' list which I and others have described a few times on this forum, and which should be discussed in any decent prep book; if no one else explains it here, I will when I have a bit more free time.