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
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-499
The sum of the first 500 positive integers is: 1 + 2 + 3 + ... + 498 + 499 + 500 = 125250
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.
All integers from 1 to 200.
20,100