Session 1

1. Explain to students that tukutuku panels are made from crossed weaving patterns.
   Here is a sequence of the first four triangular (tapatoru) number patterns.
   Either provide Copymaster One or get students to build the designs with square tiles.
   
   How many squares make up each term in the sequence?
   Is there a pattern that allows you to predict the number of squares in the fifth pattern?

2. Build up a table of values to support students look for a pattern.

   Term (Tn)	1	2	3	4
   Number of squares	1	3	6	10

3. Using the table, work together as a class to predict terms 5 and 6. Students may develop recursive rules that focus on working with a term to get the next. For example, they may notice that the differences increase by 1, 2, 3, and 4, and anticipate this growth pattern will continue.
   Suppose you had to find the 25th term. How would you do that?
   Students should notice that the recursive method is very cumbersome for larger terms.
    
4. Ask students to add the same shapes in a different colour that have been rotated 180 degrees (shown on Copymaster 1). Two triangles form a rectangle that is easier to work with.
   
5. Discuss how to find the number of crosses in the fifth member of this sequence by looking at the developing structure. (6 rows of 5 = 30.)

6. You might add a row to the previous table:

   Term (Tn)	1	2	3	4
   Number of squares (triangles)	1	3	6	10
   Number of squares (rectangles)	2	6	12	20

   What do you notice?
   Encourage students to note that the number of squares for the triangles is half that for the rectangles.
   How can you find the size of the rectangle from the term number?
   Connecting the visual pattern with each term produces rectangles that are 1 x 2, 2 x 3, 3 x 4, 4 x 5,…Encourage students to notice that one factor is the term number and the other factor is one more.

7. Discuss what the fifth and sixth rectangular and triangular number are. (5 x 6 = 30, 30 ÷ 2 = 15 gives the fifth terms, and 6 x 7 – 42, 42 ÷ 2 = 21 gives the sixth terms.)
8. Discuss what the 1000th triangular number is. (1000 x 1001 ÷ 2 = 500 500.)
9. Generalise to the nth triangular number is. (n(n+1)/2)
10. Let T₄ stand for the 4th triangular number, and S₄ stand for the 4th square number.
11. Discuss why S₄=T₄+T₃ from the drawing. (Shown on Copymaster 1.)
12. Get students to draw/make a picture that shows why S₅=T₅+T₄.

13. Together, construct a table of values for triangular numbers using the formula n(n+1)/2.

    n	10	11	12	13	14	15	16	17
    Tn	1	3	6

14. Check from this table, for a variety of values of n, that S_n=T_n+T_n-1.
15. Explain geometrically why S_n=T_n+T_n-1.

Challenge:

Show that S_n=T_n+T_n-1 by using algebra.

Session 2

Patiki (flounder) patterns.

1. Present students with this sequence of Patiki patterns. This is shown on Copymaster 2. 

   

2. Discuss how the shading helps us to recognise that in the fourth pattern that there are 4 diagonal lines of four grey crosses, and 3 diagonal lines of 3 black crosses. Therefore, there are 4 x 4 + 3 x 3 = 25 crosses altogether.
3. Have students apply this reasoning to the third pattern to see there are 3 diagonal of 3 black crosses and 2 diagonal rows of 2 grey crosses = 13 crosses.
4. Using the same reasoning, discuss how many crosses the fifth patiki pattern will need. (5 x 5 + 4 x 4 or 5² + 4²= 41).
5. Have students find the number of crosses needed for the 100^th patiki pattern. (100² + 99²= 19 801), and the nth patiki pattern (n²+(n-1)²)
6. Use another geometric method:
   The fourth patiki pattern is shaded in a new way. This is shown on Copymaster 2
   
7. Discuss how the top half shows S₄=T₄+T₃ when the grey T₃ is rotated around to fit.
8. Discuss how the bottom half shows S₃=T₃+T₂ when the black T₂ is rotated around to fit.
9. If P₄ is the number of crosses in the 4th patiki pattern discuss how all this shows P₄=S₄+S₃=4²+3²=25.
10. Discuss why P₁₀₀=S₁₀₀+S₉₉=100²+99²=19 801.
11. Draw the 5th patiki pattern using this shading and explain how this shows that the 5th patiki number is 5²+4².
12. Explain how this reasoning shows the nth patiki number is P_n=n²+(n-1)².
13. Use algebra to show that P_n=2n²-2n+1.

Session 3

Another way to find the formula for P_n is found.

1. Show the picture of the 4th patiki pattern (grey) surrounded by four T₃ shapes. This is shown on Copymaster 3.  Tiles can be used to make an S₇.
   
2. Explain how this shows P₄=S₇-4T₃.
3. Draw a picture, or use tiles that show why P₃=S₅-4T₂.
4. Complete the pattern in the formulae:
   P₃=S₅-4T₂                  P₄=S₇-4T₃                  P₅=S₉-4T₄                  P₆= ______              
   P_n= ______
   (Answer: P_n=S_2n-1-4T_n-1.)
   Hard: Use P_n=S_2n-1-4T_n-1 to show:
   

Session 4

1. A string of 5 patterns based on the 3rd triangular number is shown (Copymaster 4). Shading has been added to assist working the number of squares in the pattern.

   

2. Let students find a way to structure the figure to count the number of squares. Students may notice that the pattern is made of overlapping triangular numbers. The overlaps mean that 6 x T₄ is more than the total of squares. There are four overlaps.
3. Students might also recognise that the pattern begins with T₄ and adds on five squares for each additional section.
   Discuss why the total number of squares for Term 5 (Five sections) is both:
   6 + 4 x 5 = 26 and 5 x 6 - 4 = 26.
4. If this pattern were repeated to 1000 sections what is the number of squares needed? (6 + 999 x 5 or 1000 x 5 + 1.)
5. If this pattern were repeated m times what is the number of squares needed? (16+5(m-1) or 5m+1.)

6. Show a string of four triangular patterns (shown on Copymaster 4).

   

7. How many squares are needed to make the 100th pattern?
   (Likely answers: 10 + 99 x 9 or 100 x 9
8. How many squares are needed to make the mth pattern?
9. (Likely answers: 10 + 9(m - 1) or 9(m + 1.)
10. Develop or present a general rule:
    If the nth triangular pattern is repeated m times how many squares are needed?
    

Session 5

A string of six patiki patterns is shown (Copymaster 5). It is based on repeating the 3rd patiki pattern. Shading has been added to assist working the number of squares in the pattern.

1. Knowing that P₃=3²+2²=13 discuss why the number of squares is 13 + 5 x 12 or
   6 x 12 + 1.
2. If this patiki pattern were repeated 100 times what is the number of squares needed?
   (13 + 99 x 12 or 100 x 12 + 1 which is 1201.)
3. If this patiki pattern were repeated m times what is the number of squares needed?
   (13+12(m-1) and 12m+1.)
4. Are 13+12(m-1) and 12m+1 the same? Demonstrate that they are the same by algebra.
5. Repeat for a string of four patiki patterns (Copymaster 5) This pattern is based on repeating the fourth patiki pattern.
   
6. Find the number of squares for the 100th pattern. (P₄=4²+3²=25, so the 100th pattern needs 25 + 99 x 24 or 100 x 24 + 1 = 2401.)
7. Find the number of squares for the mth pattern. (P₄=4²+3²=25, so the number of squares is 25+24(m-1) or 24m+1.)
8. Repeat for patterns based on other sizes of patiki patterns.

Getting started

We start off with an activity that should be straightforward for most students. Then we look at another where the pattern is clear but may be harder to state precisely in words. In addition to getting the students to put the patterns into words here we also look at some basic properties of even and odd numbers. These properties are that any even number plus any even number is even; any odd number plus any odd number is even; and any even number plus any odd number is odd.

1. Build the red ‘factories’ in the diagram (see also Copymaster 1) below using square pieces of paper or magnetic tiles on a whiteboard.  

   

2. Ask questions such as the following:
   How many red squares do we need for the first red factory? The second? The third?
   How many red squares will we need for the tenth factory? The hundredth? The 124th?
   What can you say about the number of squares that you will need for any factory in this sequence? (They’re always even. )
3. Explain what is meant by the general rule for a pattern. That it’s the way of finding the number associated with a given term. So that if you know the term of the sequence you can find the number associated with that term. For instance, if we knew the general rule for the factory pattern above, we could find the number of squares that the sixth factory, say, has or the tenth factory, or whatever factory number we liked.
   What do you think the general rule for the pattern for these red factories is? (Twice the number of the term. )
   Can we test this out with a few examples? Does it work for the first factory? The second factory? The tenth factory?
4. While we’re looking at this pattern, what happens if we put two of the red factories together?
   Let’s experiment. Put the second and third factories together.
   What do we get? What is the significance of the 5 here?

   

5. If we add any red factory to any red factory will we always get another red factory?
   Experiment some more to see if this is true or not.
   If it is what is the number of the answer factory? How can you find out?
6. What does this tell you about adding an even number to another even number?

   EVEN + EVEN = EVEN

7. Now I want you to look at these green factories (see Copymaster 2). Answer as many of the questions as you can. (The general rule here is twice the number of the term before plus 1. )
8. When they have had an appropriate length of time on that task, get them together to answer questions similar to those below. First emphasise the general rule. (Here you have to double the number of the term and subtract one. This pattern is clearly the odd numbers but it may not be easy to say what the general rule is in words. ) Second emphasise the fact that even + even = even, odd + odd = even and even + odd = odd. Show using the red and green factories how this comes about.

Exploring

The students now work at various stations continuing different number patterns. The patterns explored here are powers of a number and the ‘figurate’ numbers (such as the square numbers, the pentagonal numbers and the hexagonal numbers.

The station work will take about three days. On each day get the whole class together at an appropriate time to discuss the results of their work. Check that they have been able to answer all of the questions correctly and understand what they have been doing.

You may need to remind students that if they are stuck then a good way to proceed is to make a table. Place special emphasis on the general rule that exists in each pattern. Tell them to be careful because sometimes there is more than one pattern to be found.

Day 1

The material for these stations is on Copymasters 3 and 4.The students should answer the questions posed on the Copymasters. These two stations explore the shape numbers, where the shapes are the square and the pentagon.

In this piece of work, pay special attention to the ability to put the general rule into words. This shouldn’t be too difficult. In the first case the numbers are 4, 8, 12, 16, … and so the general rule is four times the number of the term. In the second case the numbers are 5, 10, 15, 20, … and so the general rule is five times the number of the term.

Day 2

The material for these stations is on Copymasters 5 and 6. The children should answer various questions posed on the Copymasters.
On the first of these Copymasters we have another shape number - this time the hexagon. So the numbers are 6, 12, 18, 24. The general rule is that the pattern number is 6 times the term number.
On the second of these Copymasters we look at a pattern that is increasing by a multiple that is a fraction. Here the numbers are 8, 12, 18, 27, 81/2, … The general rule is to multiply the previous term number by 3/2. However, this pattern can be continued another way. The students could get 8, 12, 18, 26, … Here the numbers are increasing by 4, 6, 8, … So in this case it is easier to look at the recurrence rule. Finding the general rule is quite hard.

Day 3

The material for these stations is on Copymasters 7 and 8. Again the students complete questions that are on the Copymasters.
Here again concentrate on the general rule. In these patterns, there are at least two rules that they may find here. One is easier to tackle using the general rule and the other is easier with the recurrence rule. Encourage them to find both rules. However, they may find it hard to describe the second of these rules.
In Copymaster 7 the patterns are 1, 2, 4, 8, 16, … and 1, 2, 4, 7, 11, … The first pattern is the powers of 2 (general rule: the number of the pattern is 2 raised to the power of the term minus one). In the second pattern the difference between consecutive terms is the pattern 1, 2, 3, 4, … The increase between two terms is whole number of the first of the two terms (the recurrence rule).
In Copymaster 8 the patterns are 1, 3, 9, 27, 81, … and 1, 3, 9, 19, 33, … The first pattern is the powers of 3 (general rule: the number of the pattern is 3 raised to the power of the term minus one). In the second pattern the difference between consecutive terms is the pattern 2, 6, 10, 14, … The increase between two terms starts at 2 and increases by 4 more between each consecutive pair of terms (the recurrence rule). This second rule is much harder to describe.

Reflecting

On the final day let the class make up their own building patterns. They should concentrate on making one building pattern that has a simple general rule (like the shape patterns) and one building that has two possible patterns. They should give either the general rule or the recurrence rule, whichever is the simplest.

Bring the class together and discuss some of the patterns that they have made. Concentrate on the general rule for those patterns. Try to get all of the students to put the general rule into words. Ask them to do that.