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Pairing up with Gauss

August 25, 2007

[A number of people seem to be finding this post by searching for things like “adding up numbers in a sequence” or “what is the sum of all the numbers from…” or “summing arithmetic sequences” etc. I hope that you will stay and read this article, even though it doesn’t give you the formula. It walks through the method of finding such sums at a level appropriate for about 4th grade and up, and the derivation of the formula at a level appropriate for middle schoolers. Once you’ve learned to derive the formula, you need not memorize it, since you can re-derive it as needed.]

Here is of a lesson I developed for group of 10- to 12-year-olds. I plan to use it the first time I meet with them this year. I will first ask them to think about how they might add up all the numbers from 1 to 100, and we’ll talk about the different ways it can be done, then talk about the Gauss story, and play with lots of different kinds of sequences. I have shown many of them the Gauss trick before when playing with triangle numbers, etc. But I’m not sure how many will remember it, and in any case, this lesson is designed to take them to the next level.

They have had only very limited exposure to algebraic notation and variables, so I’m pushing things at the end, but we’ll see how it goes. I may spread this out over two class periods; I don’t think one will give them time to play with the numbers and try playing with the variables, but I’m never sure how long things will take, so again, we’ll see how it goes…

Here’s a nice illustration of some other ways of adding up sequences like this. (Full article here.) I’m hoping that my group will come up with some of these during our initial discussion of how we might approach adding up these numbers.

Young Gauss’s “trick” for finding the sum of an arithmetic progression is usually explained in terms of adding pairs of elements from opposite ends of the sequence, so that all the pairs have the same sum. One way to envision this process is to fold the series in half with a hairpin bend. Another approach is to write the series twice, once in ascending and once in descending order. A third method selects just a single pair of elements, typically the first and last, in order the calculate the average. Finally, some tellers of the story point out that the formula for summing the first n natural numbers also generates the nth triangular number; in effect, the sum is half the area of an n-by-n+1 rectangle. Brian Hayes

Adding Arithmetic Sequences by Pairing Off

Legend has it that when the great mathematician Carl Gauss was a young boy, his teacher asked him to add all the numbers from 1 to 100. Gauss quickly realized that there was a fast way of doing this, paired numbers from each end, and multiplied by the number of pairs.

sum to 100 jpeg

We can see that this sequence contains several pairs, each of which adds up to 101.

Now all we need to do is figure out how many pairs there are. Since there are 100 numbers, and there are 2 numbers in each pair, there are 50 pairs. So there are 50 101’s to add up. When we add 101 50 times, we get 50 × 101 = 5050. So…

1 + 2 + 3 + … + 98 + 99 + 100 = 5050.

Pairing Up – Practice

Use the pairing up method to find each of the following sums:

1 + 2 + 3 + … + 28 + 29 + 30
How many numbers:_____
How many pairs:_____
Sum of each pair:_____
Overall sum:_____

1 + 2 + 3 + … + 48 + 49 + 50
How many numbers:_____
How many pairs:_____
Sum of each pair:_____
Overall sum:_____

1 + 2 + 3 + … + 23 + 24 + 25
How many numbers:_____
How many pairs:_____
Sum of each pair:_____
Overall sum:_____
Use this sequence to explain why the same method even works for sequences containing odd numbers of entries.

(I’m going to cut out all of the “routine” questions for the rest here, but they’re all on the worksheet for the kids.)

2 + 4 + 6 + … + 36 + 38 + 40
11 + 12 + 13 + … + 48 + 49 + 50
5 + 6 + 7 + … + 63 + 64 + 65

1 + 4 + 7 + … + 64 + 67 + 70
How many numbers*:_____
*Careful, this one’s tricky. See if you can figure this out by finding a pattern and a rule. Explain what you did.

Pairing up – Finding the Rules

1 + 2 + 3 + … + (n-2) + (n-1) + n
How many numbers:_____
How many pairs:_____
Sum of each pair:_____
Overall sum:________

2 + 4 + 6 + … + (2n-4) + (2n-2) + 2n
How many numbers:_____
How many pairs:_____
Sum of each pair:_____
Overall sum:________

a + (a+d) + (a+2d) + … + (a+(n-3)d) + (a+(n-2)d) + (a+(n-1)d)
Fill in this description: This is a sequence of _____ numbers starting from _____ and increasing by _____ each time.
How many numbers:_____
How many pairs:_____
Sum of each pair:_____
Overall sum:________

Here’s a link to a copy of my worksheet. Feel free to use it, as long as you retain my copyright notice at the bottom of each page. If you find it useful, I’d love it if you’d let me know.

Pairing Off With Gauss

Related Post: Visit To Guangyang Primary School (relates a 4th grade lesson on this topic — scroll way down)

Follow-ups:

Pairing Up with Gauss — Follow-up I

Pairing Up with Gauss — Follow-up II

53 Comments leave one →
  1. August 27, 2007 9:46 am

    Someone found my blog yesterday with a search for “sum of first 50 multiples of 11”. I hope this post helped them figure it out!

  2. Clueless permalink
    August 27, 2007 3:16 pm

    Hi Mathmom,

    In your practice exercises, you may want to give them a decreasing sequence also. Also you may want to illustrate exercises which involve 100 terms or so, and thus make the point of infeasibility via brute force.

    TC

  3. August 27, 2007 6:57 pm

    Thanks TC, those are good suggestions.

  4. August 29, 2007 6:43 am

    Interestingly, A Learning Computation post on Social Rank led me to this post where the author uses the formula for the sum of the first n natural numbers as an example of using math to write more efficient computer programs. It might be natural to write a computer program to find the sum of the first n natural numbers to write a loop to add them one by one, but as Talat points out, it is just as inefficient for a computer (though considerably faster) to do that as for a human.

  5. September 15, 2007 1:10 pm

    See this post for a follow-up on my presentation of the first part of this lesson to my middle school group.

  6. October 4, 2007 8:46 pm

    hi
    the worksheet URL returns a 404 error.

    :0(

  7. October 4, 2007 9:02 pm

    Indeed it is. It used to work, and I can’t see why it shouldn’t still be working. I put a question up on the support forum to see if I can get it straightened out.

  8. October 5, 2007 9:05 am

    The link should be working again now. Thanks for the heads up, Andree.

  9. Hennerz permalink
    October 8, 2007 11:03 am

    Cool way for my boring maths prep!!!

  10. Kds permalink
    December 6, 2007 5:28 am

    This was discovered by Ramanujan in India, while he was in 4th Grade School !

  11. December 6, 2007 11:16 am

    Kds, I’m not sure what your point is. It was certainly a well-understood method before Ramanujan was in the 4th grade, and I don’t think it’s particularly remarkable for a 4th grader to understand it. (I show it to kids that age all the time.) Among all the things that are impressive about that mathematician, I would say that even discovering this method independently in 4th grade would not be anywhere near the top of the list. 😉

  12. Genielinda permalink
    August 23, 2008 3:02 pm

    I must tell you your explanation made this simple for this old grandma who was trying to help out her 12 year old 6th grader! Have to love Google. It made perfect sense to me, although I’ll never have to do it again. ha ha

  13. Eric Schwing permalink
    October 13, 2008 6:40 am

    Read Music of the Primes. It seems partitioning has to do with additions or at least factoring down to additions. How does one use this method for Prime numbers or can it be done?

    Also how did Euler’s product 1 + 1/4+1/9+1/16 zero in to 8/5?
    I was always challanged with high math and now that I am 59 cannot figure out how this equation came to 8/5?

    Also how did Euler use this equation to link it to quadrature of the circle
    which if I understand correctly deals with squaring a circle that is impossible but does trail off to infinite sides and how does one square a circle and know it was divided by 6 for this equation

  14. Eric Schwing permalink
    October 15, 2008 7:23 pm

    ARE YOU THERE??? or is this a defunct site.??

  15. October 15, 2008 7:54 pm

    Hi Eric,

    Thanks for you comments and sorry for the slow response. It’s an extremely busy time of year for me right now…

    The series 1 + 1/4 + 1/9 + … does not sum to 8/5 but rather to (pi^2)/6, which is approximately 8/5. As to how that relates to squaring the circle, I’m not sure offhand.

  16. Eric Schwing permalink
    October 16, 2008 7:48 pm

    Mathmom; thanks for the response….how did they arive close to 8/5 ??

  17. Eric Schwing permalink
    October 16, 2008 7:50 pm

    oops…arrive

  18. October 16, 2008 8:07 pm

    well, you can arrive at a “close” approximation of the sum of the series by figuring out the partial sums:

    S0 = 1
    S1 = 1 + 1/4 = 1.25
    S2 = 1 + 1/4 + 1/9 = 1.36111…
    S3 = 1 + 1/4 + 1/9 + 1/16 = 1.4236111…
    S4 = 1 + 1/4 + 1/9 + 1/16 + 1/25 = 1.46361111

    pi^2/6 is 1.64493407 and 8/5 is 1.6 so it will be a while before those partial sums start looking really close. Using an excel spreadsheet, I found that S30 = 1.613191. So if you just want an approximation, you can do it that way.

    To prove that it converges to pi^2/6 requires calculus. It’s way beyond the scope of this blog (wherein I discuss primarily K-8 math education) and also an area where I’m pretty rusty these days, but click the link above if you want to see one way of proving it.

  19. Eric Schwing permalink
    October 20, 2008 5:03 pm

    Wow, thankyou..

  20. Eric Schwing permalink
    October 20, 2008 5:39 pm

    One more question. I can see that each series increase increases slightly.
    but how are these complex numbers greater than 1 reconstructed??

    So for example in your third series what is the complex number reconstructed and how is it done. It is easy to see that 5 by 8 is 1.6
    but how are each increases reconstructed to that convergent new complex number

    And finally does the series of 30 terms get so small that it does in fact converge to that 8/5 in other words each series has their own converging complex number and how did they know to stop there?

  21. October 20, 2008 5:47 pm

    I don’t understand what you are asking about reconstructing complex numbers. And, for starters, I assume you’re not talking about complex numbers in the mathematical sense. But which numbers are you asking about “reconstructing”?

    If you are asking how I got the different S1, S2, S3, …. sums it is just by adding fractions (although the result is expressed in decimal notation):

    S2 = 1 + 1/4 + 1/9 = 1 + 9/36 + 4/36 = 1 + 13/36 = 1.36111… in decimal notation

    There is only one series here, and it is infinite, and it does converge, but not to 8/5 but rather to pi^2/6, which is somewhat close to 8/5. The only reason I can think of that you ever saw 8/5 associated with this series would have been as a rough estimate, though I don’t know where it came from or why someone felt it would be valuable to present that estimate, since it is not that close to the true value.

  22. Eric Schwing permalink
    October 20, 2008 5:53 pm

    also, the second term 1.25 is the common denominator 5 and is not an infinite complex number…Is it because even numbers are all finite and odd numbers produce infinte series numbers?

    Can you recommend a book that helps me do a crash course on abstract relations in math? I am a History buff and social theorist and one of the reasons I did not do so well in math, It was taught as rote calculations without an overall view to connections, just like History is taught in school.

  23. October 20, 2008 6:25 pm

    Some fractions, such as 1/3 are repeating decimals in decimal notation. They are not complex numbers, they are rational numbers. It is not a matter of even or odd, for example 3/5 = .6

    I don’t know of a book offhand, but you might want to check out the online course discussed here. I’ve also heard good things about the videos available from The Teaching Company, but they’re fairly expensive.

    Ask Dr. Math is a great online resource for learning more about math topics.

    Hope that helps.

  24. December 2, 2008 4:44 pm

    Wow! you explain this better than my college professor in Intermediate Algebra. If kids can learn this stuff early in life, more power to them. It stinks when an adult brain tries to learn something, we become clueless. Great blog.

  25. December 2, 2008 6:26 pm

    Rebecca,

    Thanks for the kind words!

  26. Harry permalink
    January 23, 2009 6:51 am

    Hi

    I have a team of 8 people who need to be paired up in 2’s within pairings of 4’s over 4 days.
    eg: Day1: 1&2 plays with 3&4
    Day2: 1&5 plays with 2$6
    etc
    What is the best formula to divide them up evenly so that everyone gets to play with everyone else on different days in different 4balls.
    I know some will play more with others as they are only 8 people, but I am looking for the best fit.
    Please could you gelp me with some form of algorithm or something.

    • November 25, 2011 2:53 pm

      While I have always enjoyed hearing this story about Guass; the problems seem to miss the point of how one comes to a general solution to this class of problem. To show what I mean try aplying your excercise summing lists of numbers which are odd. If you take the sum 1+2+3+4+5 and try to group the numbers in pairs you will find yourself in a bind. The formula for these sorts of problems n(n+1)/2 can be derived by adding your list (which starts with one) 1+2+3+4+5 to itself 5+4+3+2+1 which will yeild 6+6+6+6+6=30 divide by two and you have the correct solution to the problem.

      Forgive me for being pendantic.

  27. laurie permalink
    January 12, 2012 5:20 pm

    so how do u get! am lost
    1 + 2 + 3 =…+ 400

  28. January 12, 2012 7:54 pm

    laurie, pair them up. You will see that if you take the first and last, they sum to 401 (1+400). And then the 2nd and next-to-last also sum to 401 (2+399) and so on. Once you figure out how many pairs you have, you just multiply that by the pair sum (in this case 401) to get the sum of the series.

    • Mary permalink
      September 21, 2012 12:56 am

      hey

      could you pls help me, i don’t understand how to find the sum of this ;
      1+1\9+1\25+1|49……..

      pls help me

  29. Wagh harshal keshav permalink
    January 25, 2012 7:56 am

    I hv lookd these all technics and also take their applications, realy its very better way for young1s solve these type of series . Only suggetion is that to hv the formula for finding the sum of factorials .

  30. Wagh harshal keshav permalink
    January 25, 2012 8:00 am

    I hv lookd these all technics and also take their applications, realy its very better way for young1s solve these type of series . Only suggetion is that to hv the formula for finding the sum of factorials .otherwise overall best

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  36. July 14, 2013 11:36 pm

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  37. Pooja permalink
    January 4, 2014 8:23 am

    hey! why does the pairing method only work on arithmetic sequences??

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Trackbacks

  1. Pairing Up with Gauss — Follow-up I « Ramblings of a Math Mom
  2. Pairing Up with Gauss — Follow-up II « Ramblings of a Math Mom
  3. Elementary Problem Solving « Ramblings of a Math Mom
  4. 20th Carnival of Mathematics « Ramblings of a Math Mom
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