tags:

231 + 237 + … + 849 + 855

I know what the sum is, but I am curious how you would find it. And moreso, how you might teach it.

Incomplete background: I’ve taught this one way (A) for some time. I have always noticed some kids doing something different (B); it (B) made sense, but I didn’t teach it. Last year I let the kids explain the alternate method (B), although it remained clear that I considered (A) primary and (B) alternate. This year (B) showed up again (persistent!) and I decided to have the kids who did it explain again, but this time I tried to value (B) for some advantages it might have, and (A) for some advantages it might have.

And I find that the kids still prefer ‘my’ way, but only by about 60-40. Before I tell how I do it and what the other method is, I’d be curious to find out ho you, readers, attack this one.

Thanks!

1. October 22, 2008 pm31 11:25 pm 11:25 pm

I’ve written a series of blog posts about how I teach finding the sum of an arithmetic series. The tricky part about your example is finding how many elements are in the series. Kids are usually off by one. I usually have them try their “method” for a smaller series with 2 or 3 or 4 elements, to see where they’re missing an element.

October 22, 2008 pm31 11:27 pm 11:27 pm

I started with a smaller sum: 231 + 237 + 243 + 255. I figured out that that was 231(4) + 6(3 + 2 + 1)

The 4 was from the original 4 terms, and the 3 + 2 + 1 was from the successive multiples of 6 being added to each term after 231. So for 231 + 237 + ….. + 849 + 855, I figured there are 105 terms. That would make 231(105) + 6(104 + 103 + 102 + 101 + …… + 2 + 1), which equals 57,015.

Then I checked my work in Excel. :-)

3. October 22, 2008 pm31 11:27 pm 11:27 pm

oops, the firs link above should be just this without the comment part on the end, but you can just scroll back to the top of the page if you use that one.

October 22, 2008 pm31 11:28 pm 11:28 pm

Now I’m curious to see how you teach it (both ways!)

5. October 23, 2008 am31 12:44 am 12:44 am

http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning

October 23, 2008 am31 3:01 am 3:01 am

My first thought would be to subtract 231 from all of them, getting the sum 0+6+12+…+624.

Starting at 0 might be unpleasant, so in hindsight let’s just subtract 225 from each, leaving 6+12+18+…+630.

Now factoring out 6 from the sum leaves 6 (1+2+3+…+105), and now we know that there were 105 summands, that we subtracted a total of 105 x 225, and that what was left was equal to 6 x (105×106/2).

(Yes, I’ve taught these this way, and in my experience [college] students can manage this. Once they get the hang of it, they tend to be better at this than any attempt to use the formula for the nth term of an arithmetic progression. But your mileage may vary.)

7. October 23, 2008 am31 4:34 am 4:34 am

I’d say this sequence has (855 – 231)/6 + 1 = 624/6 + 1 = 105 terms, so the sum is 0.5*105*(231+855) = 105*1086/2 = 105*543/2.

8. October 23, 2008 am31 5:37 am 5:37 am

Mathmom – “If I assigned problems 12 – 17, how many problems would that be?” That’s how I start this course, and we never stop referring back to it. Crucial for counting. Endpoints. Posts vs rails. They have to be forced to think constantly about this.

My method, (A) is almost exactly what TwoPi presents. It reminds me of working with integrals… really. I keep sigma notation in place, which few kids like, and then I do it again with dot dot dot lists, which they like better

(dot dot dot list is, for example, 224 + 235 + … (dot dot dot) + 345 + 356)

The kids bring up something like Alon’s. They find the average (average of first and last works fine) and multiply by the number of numbers, in this case subtract 225 and divide by 6. So (B) is n*avg, and it works fine.

9. October 23, 2008 am31 5:59 am 5:59 am

I do it Alon’s way. (a.k.a. 8-year old Karl Gauss’ way.) :-) I think twopi’s is more elegant, but Alon’s is more obvious…I certainly don’t see anything wrong with getting them to understand why both work.

October 23, 2008 am31 6:01 am 6:01 am

I did this three ways:

1) Halfway between 855 and 231 is 543. find the number of terms (105) and multiply by 543. each term larger than that is canceled by one smaller than that, so it works. I think this depends on understanding averages? I wasn’t sure of this method at first so I checked with (2) and (3) below.

2) 231n +6(n*52) I know n is 105 by subtracting the 231 from 855 then adding one. So I have that many 231s, and each has six added one additional time, that’s 6*0, 6*1, 6*2, etc. added, up to 6*104. to quickly add up all those sixes, divide 104/2 then multiply by n which is basically using the average to find the sum. so it’s (105*231)+6*(105*52).

3) List them in Excel using the fill feature, then add them up with a quick =sum().The whole task took 1 minute that way. Do kids learn to use tools like Excel?

October 23, 2008 am31 6:55 am 6:55 am

Something equivalent to finding the average: write the whole sum out twice, once forwards and then again, directly below it, backwards. Instead of summing across, sum down first:

231 + 237 + … + 849 + 855
855 + 849 + … + 237 + 231
==================
1086 + 1086 + … + 1086 + 1086

I think the trick in any solution method is making sure you don’t create an off-by-one error, as several people have already noted. I think TwoPi’s method is a particularly nice way of getting that part right, though “posts vs. rails” is also a nice mnemonic.

12. October 23, 2008 am31 7:50 am 7:50 am

In computer science, where this kind of off-by-one error is common (in setting up the bounds for a loop mostly) it is sometimes called a “fencepost error”.

13. October 23, 2008 pm31 3:57 pm 3:57 pm

I read, and heartily recommend, Jason’s American Scientist article, above.

(there’s also a nice forensic aspect, as the article tries to establish where and when different details were added to the Gauss story/legend)

14. October 23, 2008 pm31 5:08 pm 5:08 pm

The article Jason linked is my favorite on the topic as well.

15. October 23, 2008 pm31 7:00 pm 7:00 pm

Simple case of AP as I see it. The no. of terms are 105. After finding this out. We can find the sum for 105 terms with the first term as 231 and the common difference as 6.

The sum S = (105) * (543) = 57,015.

I still didn’t get what the complication in this this was…

16. October 23, 2008 pm31 11:38 pm 11:38 pm

I teach the same “a+L” method that many of your kids like, and would regard it as the standard method. It is (to my mind) more elegant and intuitive than yours.

How do you prove that the n integers are $\frac{1}{2}n(n+1)$ without using that idea anyway?

17. October 29, 2008 am31 2:04 am 2:04 am

Dr. Rick, if you check the article I linked, one clever alternate method is through geometry. Just take an n x (n+1) rectangle and chop it in half on the diagonal, and by the rows you’ll get the numbers 1 through n.

18. October 30, 2008 am31 3:10 am 3:10 am

Jason – yes, but that *is* the same idea :). (That the terms counting in from opposite ends always sum to a+L, which is certainly how I teach it.)