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 *n*th
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 15^{2}
= **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 *n*th
term of such a progression, namely *a*_{n}_{ }=
*a*_{1} + (*n* – 1)*d,* where *a*_{n} is
the *n*th term, *a*_{1} 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 *n*th term
of the progression could be found with the formula *a*_{n}_{ }= *a*_{1} × *r*^{n }^{– 1} where *a*_{n}_{ }is the *n*th term, *a*_{1} is the first term, and *r* is the common ratio (in our example, five). So the fifteenth term
would be 2 × 5^{14}
(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^{3 }= 64 , and 4 + 64 = 68,
we would have 5, 5^{3 }= **125**,
and 5 + 125 = **130**;
**6**, **216** (6^{3}), **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!”

Ralph 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.