For any positive integer n, the sum of the first n positive integers equals (n(n+1))/2. what is teh sum of all the even integers between 99 and 301.

The correct answer in the OG is 20,200. The OG solves this problem by plugging 159 and 49 as n because " the sum of the even integers between 99 and 301 is the sum of the even integers from 100 through 300 or the sum of the 50th even integer through the 150 even integers. However, when solving the problem they multiply each of the equation by two: 2((150(150+1))/2) - 2((49(49+1)/2).

For the life of me i can't understand why they multiply the two formula by two. Can anyone explain?

They're conceptually breaking this into two pieces --
We have
100 + 102 + 104 + 106 + ... + 298 + 300
This can be expressed as
2*50 + 2*51 + 2* 52 * 2*53 * ... + 2*149 + 2*150
We can factor a 2 out of this:
2 * (10 + 51 + 52 + 53 + ... + 149 + 150)

Then, they perform the sum on the part in the parentheses and then multiply by 2 because of the factored out 2.

If this doesn't answer your question, though, let me know,

Veritas Help

How do you get from 2(50+51+52.......150) to 20,200? Is there a short cut for adding consecutive numbers within the parentheses?

Yep - there's a shortcut.

Let's look at a shorter example first:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

Look at the first and last term -- 1+10 = 11
Then, look at the second and second-to-last term: 2 + 9 = 11
The same will apply all the way in.
We have a total of 10 terms. This is an even number, so we have 5 pairs. Each pair will total 11.
Our sum will be 55.

In the bigger scenario:
50 + 51 + 51 + ... + 149 + 150
We have a total of 101 terms, so we have 50 pairs, plus 1 extra term in the middle
Each pair (starting at end points) will total 200. The middle term will be 100
So, we have 50*200 + 100 = 10100 as the sum of the terms from 50 to 150...

Hope this helps -

Veritas Help