For any positive integer n, the sum of

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

2 Explanations

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Sam Kinsman

Hi Yeshna,

I think that Mike's solution is the most straightforward. However, if you have a hard time following that, and would like to use the formula that's given in the problem, you could do the following:

The sum of the first n positive integers equals n(n+1)/2

Find the sum of all even integers from 1 to 301. We have (2+4+6+....+300). We can factor out a 2 here, to get 2 * (1+2+...+150). Using the formula in the problem, the sum of (1+2+...+150) is (150*151)/2 = 75*151. So we have 2*75*151 = 150*151

Next, find the sum of all even integers from 1 to 99. We have (2+4+6+....+98). We can factor out a 2 here, to get 2 * (1+2+...+49). Using the formula in the problem, the sum of (1+2+...+49) is (49*50)/2 = 49*25. So we have 2*49*25 = 49*50

Now, we need to take the sum of all even integers from 1 to 301, and subtract the sum of all even integers from 1 to 99. So we have:

(the sum of all even integers from 1 to 301) - (the sum of all even integers from 1 to 99)

This becomes (150*151) - (49*50) = 50*(453 - 49) = 50 * 404 = 20200

This is an alternative way to solve this problem - but it's not necessarily easier! This is a tricky problem, so there isn't really an easy way to solve it. If this method is hard to follow, you may want to re-watch Mike's explanation video! :)

Nov 18, 2017 • Comment

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Mike McGarry, Magoosh Tutor