Recently, inquiring readers six-year old (!) Julien and his Mom, asked about Triangular Numbers:
Hello, My son is currently working his way through The Number Devil (by Hans Magnus Enzenberger) and is enjoying it thoroughly. He was particularly happy about the triangular numbers in Chapter 5 because I had just coincidentally given him a worksheet which involved calculating how many blocks would be required to complete a series of steps from 1 all the way up to 20. (1+2+3+4+5+6+7…)
He made the connection between his worksheet and the triangular numbers and tried out the (much more efficient) trick the number devil provides (p.101) for solving such a problem:
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1
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2
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3
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4
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5
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6
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7
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8
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9
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10
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+20
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+19
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+18
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+17
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+15
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+18
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+14
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+13
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+12
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+11
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21
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21
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21
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21
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21
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21
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21
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21
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21
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21
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21*10=210 However, he immediately realized there will be a problem if you have an odd number of steps.
Can this method be used with an odd number, and if so, how would it work? We have read your page on triangular numbers and used the equation you provide (x2+x)/2 and that works great. But, he (and I) just wonder if we are missing something about the method discussed in the book (because the number devil doesn’t say it’s ONLY for even numbers).
Thank you for your time!
Professor Homunculus replies:
Hi Julien and Miram,
That’s a great question, and I’m very happy that you are interested in things like this. It is the sign of a mind that knows how to have some fun with patterns. And it is a great inspiration to folks who wonder at what age kids can get interested in math.
I haven’t got a copy of The Numbers Devil anymore, and it’s been a long time since I’ve read it, but it still remains one of my favorite books to introduce real math to people with.
I hope I’m understanding your question correctly. If I am, the following should clear it up for you and your son. There are several ways of looking at this pattern to make it “work” for odds and even numbers. There are probably many more than I will ever know, but here is one of the ways that I like to view it:
Include zero in the mix. It doesn’t change the sum of the number, but it makes the same pattern as above. For example, if you wanted to add the series from 1 to 5: Just keep in mind that adding the series from 1 to 5 is the same as adding the series from 0 to 5.
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0
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1
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2
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+5
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+4
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+3
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5
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5
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5
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5 * 3 =15 It works! Let’s try it with 19:
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0
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1
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4
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5
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6
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9
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+19
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+18
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+17
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+16
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+15
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+14
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+13
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+12
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+11
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+10
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19
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19
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19
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19
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19
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19
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19
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19
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19
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19
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19* 10 = 190 It works again!
Is that what you were getting at? Did this help solve your son’s dilemma? Please let me know if it did.
All the best, Brian (a.k.a. Professor Homunculus at MathMojo.com )
BTW, readers not familiar with the formula given above ( (x2+x)/2 ) can find out more about it and triangular numbers in general at the MathMojo.com page at Adding Triangular Numbers
A challenge:
Can any reader see a mental shortcut of how to arrive at the sum of a series from 1 to n? There may be more than one. (Hint: The ones I use are slightly different for even and odd numbers).
I read our library’s copy of The Number Devil a couple of years ago with our oldest daughter – it was fun and interesting! Thank you for bringing up this subject.
The standard way to do this (seen in many high school math texts) is to write all the numbers in the sum in increasing order, and under that write down all the numbers in the sum in decreasing order. If n = 20, as in the problem given, each column adds to 21 and there are 20 columns, so the total is 420. Since we wrote the sum down twice, the sum of the of the numbers is 420/2 = 210.
I tell my students this is just like counting the number of people in a crowd by counting the total number of legs and dividing by two.
Peter,
Right on the money. This is also the way that is described on this Mathmojo page about triangular numbers.
for series 1,2,3, ….. ,n
1) sum of series = (n*(n+1))/2
eg from 1 to 19,
sum = (19*20)/2 = 190
The “diagram” below shows the reasoning behind this for the series 1,2,3,4. The ‘*’ represents the value of numbers 1,2,3 and 4. To find the number of *s, add an equal number of ’0′s to form a rectangle. This will form a 4 x 5 rectangle. The number of *s is given by (4*5)/2
12345
1 *0000
2 **000
3 ***00
4 ****0
2) sum = ((1+n)/2) * n
sum = ( (1+19)/2 ) * 19 = 190
This finds the average of the series (1+n)/2 and multiply the average by the number of terms in the series. Average of 1 to 19 is (1+19)/2 or 10. Multiply by 19 to get the sum of 190.
3) Multiply the median number of the series by the number of terms in the series. This is somewhat similar to (2) above.
If n is odd, the median number is (n+1)/2. The median number for 1 to 19 is (19+1)/2 or 10. Sum of 1 to 19 is 10 * 19 = 190.
If n is even (eg 1 to 20), there is no exact median number. Therefore to get the sum of 1 to 20, first get the sum of 1 to 19 (as above) then add the last number 20 to get the final sum. sum of 1 to 20 is 190 + 20 = 210
Hey, Thanks for the excellent explanation. I appreciate it, and I’m sure other readers will, too.