The formula n*(n+1) is used to find the sum of n positive integers. Th sum of positive integers up to 500 can be calculated as 250*251=62,750.
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Oh, dude, that's like asking me to count all the hairs on my head. Okay, so you want the sum of all even numbers from 2 to 500? That's like adding 2 + 4 + 6... all the way up to 500. But like, who has time for that? The sum is 125,250. Happy now?
It is 1.
501
Adding all integers from 33 to 112 inclusive gives you 5800.
There's a trick to this one. Let 1 + 2 + ... + 500 be the sum of all integers 1 to 500, called X Now, imagine we want to add the sum the integers from 1 to 500 and to the sum of the integers from 1 to 500. This would give us 2*X we can write it as 1 + 2 + ... + 500 + 1 + 2 + ... + 500 But that's not particularly useful. What if we look at it as: 1 + 2 + ... + 499 + 500 + 500 + 499 + ... + 2 + 1 adding the numbers that are directly on top of each other, we get 500 + 1 = 501, 499 + 2 = 501... 1 + 500 = 501 Thus every term is 501, and we have 500 terms. So, we have 501*500 = 2X Thus X, the sum from 1 to 500, is (500*501)/2 = 125,250
10, 8, 6, 4, 2, 0 and all the negative even integers.