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Reflecting on the Pondering Patterns Problem

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Greetings! Over the past few months, it has been great fun sharing some of my favorite “Math Tasks to Talk About” with you and becoming a blogger in the process. The plan for the TCM blog is for a series of guest bloggers to continue adding to this rich collection as they share and discuss their favorite tasks, so I now need to step aside and make room for the next person. I hope you’ve enjoyed the tasks I’ve shared with you, and I’d certainly be delighted if you’d share your thoughts by commenting on the blog!

So, how did you and your students respond to the Pondering Patterns task? Let’s start by looking at the four patterns talked about in the first paragraph and how students might go about finding a specified term in each pattern. The example I used was finding the fifteenth term, so let’s begin with that. The first pattern was the one generated by the Handshake problem: 1, 3, 6, 10, 15, 21, 28, . . . . Most elementary school students would likely find the fifteenth term here by just continuing the pattern out fifteen numbers: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120. But if you recall the discussion from that task, students might also recognize that the fifteenth term would be the total number of handshakes for sixteen people shaking hands, or 15 + 14 + 13 + 12 + . . . = 120, and some students might be able to generate a formula for the nth term of the pattern, n(n + 1)/2. So, for the fifteenth term, it would be (15 × 16)/2 = 240/2 = 120.

By the way, if students were sticking with “handshake reasoning,” the formula would be slightly different—the fifteenth term would be the number of handshakes for sixteen people, or (16 × 15)/2, but the result would be the same.

For the pattern generated by the  How Many Squares on a Checkerboard? Task: 1, 4, 9, 16, 25, 36, . . . , with very little prompting or scaffolding, most students would be able to recognize that for any specific term, the number would just be that term multiplied by itself, so the fifteenth term would be 152 = 225. For the arithmetic progression 1, 4, 7, 10, 13, 16, 19, again most students would just continue the pattern of adding three until they got to the fifteenth term. But, depending on the grade level and how much time the teacher wanted to spend on a discussion of arithmetic progression patterns, students could be led to determine the formula for the nth term of such a progression, namely an = a1 + (n – 1)d, where an is the nth term, a1 is the first term, and d is the common difference (in this case, three). So the fifteenth term of the progression above would be 1 + (14 × 3) = 43. Because geometric progression patterns increase in value so quickly, for elementary students, one would likely not ask for a term as large as the fifteenth term (finding the fifth or sixth term would be more appropriate), but for those of you determined to find out, in the pattern given: 2, 10, 50, 250, the nth term of the progression could be found with the formula an = a1 × rn – 1 where an is the nth term, a1 is the first term, and r is the common ratio (in our example, five). So the fifteenth term would be 2 × 514 (or a very big number!).

OK, let’s move on to the far less complicated (but perhaps more sneaky) patterns given in the last task.

Complete the following pattern:

5 -----> 4

36--> 8

11---> 1

53---> 7

942---> 14

18---> ?

49---> ?

371---> ?

This is a great example for showing how it’s possible to sometimes overthink patterns, as students and teachers try all manner of combinations of operations to get from the first number to the second one, before “stepping back” and realizing that the second number is just one less than the sum of the digits of the first number, so 18 --> 8 ; 49 --> 12 ; and 371 --> 10.

Study the numbers below, and continue the pattern by listing the next five numbers in the sequence:

1, 1, 2, 2, 8, 10, 3, 27, 30, 4, 64, 68, 5, __, __, __, __, __

This pattern becomes more obvious as you look “further in” to it, seeing 2 followed by 8; 3 followed by 27; and 4 followed by 64. The pattern is in groups of three: a number, that number cubed (raised to the third power), and then the sum of the number and the number cubed. So after 4, 4= 64 , and 4 + 64 = 68, we would have 5, 5125, and 5 + 125 = 130; 6, 216 (63), 222 as the next five numbers in the pattern.

Study the numbers below, and continue the pattern by listing the next five numbers in the sequence:

13, 4, 17, 8, 25, 7, 32, 5, 37, 10, 47, 11, 58, ___, ___, ___, ___, ___

This pattern has a starting number, followed by the number which is the sum of the digits of that number, followed by the sum of the two numbers. It continues by finding the sum of the digits of that number, adding the two again, and so on. So the next five numbers in the pattern would be 13 (sum of the digits of 58); 71 (sum of 58 and 13); 8 (sum of the digits of 71); 79 (71 + 8); 16 (sum of the digits of 79).

I hope that you and your students had some fun with these somewhat unusual patterns. Do you have a pattern to share? Any and all comments regarding these patterns and others you might want to talk about are welcome! I’ll be around to respond to your comments, and then it will be, “Ralph has left the blogosphere!” 

RalphConnellyRalph Connelly is Professor Emeritus in the Faculty of Education at Brock University in Ontario, where he taught elementary math methods courses for 30+ years. He is active in both NCTM, where he’s served on several committees, currently the Editorial Panel of TCM, and NCSM, where he’s served two terms as Canadian Director as well as on numerous committees. 

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