In this unit some Tukutuku patterns are introduced. Rotations of these patterns produce simple shapes whose area formulae are well known. From these formulae algebraic formulae for sequences can be deduced. While the unit is written to the Level 4 Achievement Objectives, the work is quite advanced and may be most suitable for students entering Level 5.
 find the number of crosses in Tukutuku panels by using areas of squares and rectangles
 find the number of crosses in repeating Tukutuku panels by using linear formulae
Triangular numbers and patiki patterns are used as a powerful way to introduce patterns leading to quadratic and linear formulae. A number of alternative geometric views are suggested. Establishing that the formulae are in fact identical is a useful application of quadratic algebra skills.
Session 5
Session 4
Session 3
Session 1
Session 2
Tukutuku panels, Patiki patterns, sequence, triangular numbers, rotation, diagonal, equations, spatial patterns, formula,
Session 1

Explain to students that Tukutuku panels are made from crossed weaving patterns.
Here is a sequence of the first four triangular number patterns. (Repeated on copymaster or get students to build them from 2 colours of tiles.) 
Ask students to add the same shapes in a different colour that have been rotated 180 degrees. (Repeated on copymaster.)

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.) Discuss what the fifth triangular number is. (30 ÷ 2 = 15.)

Discuss what the 1000th triangular number is. (1000 x 1001 ÷ 2 = 500 500.)

Discuss what the nth triangular number is. (n(n+1)/2)

Let T_{4} stand for the 4th triangular number, and S_{4} stand for the 4th square number.

Discuss why S_{4}=T_{4}+T_{3} from the drawing. (Repeated on copymaster.)

Get students to draw/make a picture that shows why S_{5}=T_{5}+T_{4}.

Construct a table of values for triangular numbers using the formula n(n+1)/2.
n
10
11
12
13
14
15
16
17
T_{n }
1
3
6

Check from this table, for a variety of values of n, that S_{n}=T_{n}+T_{n1}.

Explain geometrically why S_{n}=T_{n}+T_{n1}.
Challenge:
Show that S_{n}=T_{n}+T_{n1} by using algebra.
Session 2
Patiki (flounder) patterns.

Students should look at this sequence of Patiki patterns. (Repeated on copymaster or make out of tiles.)
 Discuss how the shading helps student recognise that in the fourth pattern that there are 4 diagonal lines of four gray crosses, and 3 diagonal lines of 3 black crosses. So there are 4 x 4 + 3 x 3 = 25 crosses altogether.
 Apply this reasoning to the third pattern to see there are 3 diagonal of 3 black crosses and 2 diagonal rows of 2 gray crosses = 13 crosses.
 Using the same reasoning discuss how many crosses the fifth patiki pattern will need. (5 x 5 + 4 x 4 or 5^{2} + 4^{2}= 41.)
 Find the number of crosses needed for the 100^{th} patiki pattern. (100^{2} + 99^{2}= 19 801), and the nth patiki pattern (n^{2}+(n1)^{2})
 Another geometric method:
The fourth patiki pattern is shaded in a new way. (Repeated on copymaster or build it from tiles.)  Discuss how the top half shows S_{4}=T_{4}+T_{3} when the grey T_{3} is rotated around to fit.
 Discuss how the bottom half shows S_{3}=T_{3}+T_{2} when the black T_{2} is rotated around to fit.
 If P_{4} is the number of crosses in the 4th patiki pattern discuss how all this shows P_{4}=S_{4}+S_{3}=4^{2}+3^{2}=25.
 Discuss why P_{100}=S_{100}+S_{99}=100^{2}+99^{2}=19 801.
 Draw the 5th patiki pattern using this shading and explain how this shows that the 5th patiki number is 5^{2}+4^{2}.
 Explain how this reasoning shows the nth patiki number is P_{n}=n^{2}+(n1)^{2}.
 Use algebra to show that P_{n}=2n^{2}2n+1.
Session 3
Yet another way to find the formula for P_{n}.
 Show the picture of the 4th patiki pattern (grey) surrounded by four T_{3} shapes. (Repeated on copymaster or make it from tiles to make an S_{7}.)
 Explain how this shows P_{4}=S_{7}4T_{3}.
 Draw a picture, or use tiles that show why P_{3}=S_{5}4T_{2}.
 Complete the pattern in the formulae:
P_{3}=S_{5}4T_{2} P_{4}=S_{7}4T_{3} P_{5}=S_{9}4T_{4} P_{6}= ______
P_{n}= ______
(Answer: P_{n}=S_{2n1}4T_{n1}.)
Hard: Use P_{n}=S_{2n1}4T_{n1} to show
Session 4
 A string of 5 patterns based on the 3rd triangular number is shown (copymaster). Shading has been added to assist working the number of squares in the pattern.

Knowing that T_{4}=6 discuss why the number of squares is
6 + 4 x 5 = 26.  If this pattern were repeated 1000 times what is the number of squares needed? (6 + 999 x 5 or 1000 x 5 + 1.)
 If this pattern were repeated m times what is the number of squares needed? (16+5(m1) or 5m+1.)
 A string of four triangular patterns is shown (copymaster).
 How many squares are neededto make the 100th pattern?
(Likely answers: 10 + 99 x 9 or 100 x 9  How many squares are needed to make the mth pattern?
 (Likely answers: 10 + 9(m  1) or 9m + 1.)
 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). It is based on repeating the 3rd patiki pattern. Shading has been added to assist working the number of squares in the pattern.
 Knowing that P_{3}=3^{2}+2^{2}=13 discuss why the number of squares is 13 + 5 x 12 or
6 x 12 + 1.  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.)  If this patiki pattern were repeated m times what is the number of squares needed?
(13+12(m1) and 12m+1.)  Are 13+12(m1) and 12m+1 the same? Demonstrate that they are the same by algebra.

Repeat for a string of four patiki patterns (copymaster) This pattern is based on repeating the fourth patiki pattern.
 Find the number of squares for the 100th pattern. (P_{4}=4^{2}+3^{2}=25, so the 100th pattern needs 25 + 99 x 24 or 100 x 24 + 1 = 2401.)
 Find the number of squares for the mth pattern. (P_{4}=4^{2}+3^{2}=25, so the number of squares is 25+24(m1) or 24m+1.)
 Repeat for patterns based on other sizes of patiki patterns.