Adding Arithmetic SequencesFrom: Norma Subject: Adding consecutive numbers How do you add the numbers from 1 to 5000 without actually doing it or using a calculator? I added 1 and 2 and got 3. 3 and 3 and got 6, but I know this takes way too long. Help! From: Doctor Pat Subject: Re: Adding consecutive numbers Norma, If you wrote out all the numbers from 1 to 5000 and then wrote them backwards underneath, you would have twice as many numbers as you needed, but the problem is easier, here's why: 1 2 3 4 ............... 4998 4999 5000 5000 4999 4998 4997 ............... 3 2 1 -------------------------------------------------------- 5001 5001 5001 5001 ............... 5001 5001 5001 Notice that if we add the two lists, we get a list that is the same number, 5001, repeating. In fact, since each of the lists is 5000 numbers long, we have, in the sums, a list of 5000 numbers that are each 5001. You have to admit that adding 5000 5001's is a lot quicker than the other way since 5000 x 5001 = 25,005,000. But wait, you say, that's too much. We were only supposed to add one list and we added two. Okay, then the answer must be half as much: 1 + 2 + 3 + ... + 5000 = 25,005,000/2 = 12,502,500 See, you can do the whole thing in your head. You can use this method if you have any number of consecutive numbers, whether they start with 1 or not. In fact, the numbers do not have to even be consecutive. They just have to be in an arithmetic sequence, that is, the difference between any two adjacent numbers must always be the same (in your example the difference was 1). Let's add up all the odd numbers from 1 to 25. We do the same thing as before: 1 3 5 ......... 21 23 25 25 23 21 ......... 5 3 1 ----------------------------------- 26 26 26 .......... 26 26 26 We have to know how many numbers there are, and in this case there are 13 twenty-sixes, so the total must be 13*26/2. In general if you have an arithmetic sequence of N numbers and you know the first and last one, you can find the sum by: Sum = N(First + Last)/2 Using this for your original set with N = 5000, first = 1 and last = 5000, we get the same result 5000(5001)/2. Hope that helps. Thanks for writing. - Doctor Pat, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/