Question 5666: Confused about problem. Need to solve this but I do not know how to use the formula. Need to know the sum of pairwise comparisons.
1+2+3+...+498+499+500
Answer by prince_abubu(198)   (Show Source):
You can put this solution on YOUR website! You might be looking for a formula that will add up the consecutive natural numbers from 1 to n. You know that f(1) = 1, f(2) = 1 + 2 = 3; f(3) = 1 + 2 + 3 = 6, etc... and ultimatley, f(n) = 1 + 2 + 3 + .... + n. But you don't want to keep on adding the consecutive terms because it'd take you forever, especially if n were big enough.
I've seen this one before and the derivation is quite clever. Look again at the expression that you wrote 1 + 2 + 3 + .... + 498 + 499 + 500. Imagine that each number were on a strip of long paper. Fold the paper exactly in half. This way, the 1 will be paired with the 500, the 2 will be paired with the 499, the 3 will be paired with the 498, and so on.
If you notice, each pair after the fold ALWAYS has a sum of 501. Aha! But how many pairs would there be? There would be 250 pairs because you folded the strip in half. So 1 + 2 + 3 + 4 + ... + 500 = 501 * 250 = 125250.
Working with our example here, N = 500, the last number in our summation. When we "paired the opposite ends", we found that the sums were 501, which really is N + 1. Each pair's sum was 501, but how many pairs were there? 250 pairs, which really is N/2.
Now, we can put together the formula  written in nicer form. As mentioned, that formula actually equals 1 + 2 + 3 + ... + n.
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