pattern talk 17
Holy cow. Students had more ways to see this pattern than any other so far. I had to extend the talk over two days because we’d already used up 20 minutes of class on the first day! I’ve been calling on students (first 3 at random) to share the equation if they have it, then I ask the class to try to see where each term in the equation comes from.
Student 1
Student 2
I love how this kid saw this!!
Student 3
Student 4
Student 1
- I see the perimeter as 4 groups of (n-1), plus the 4 corners.
- Then I see these diagonals inside [brown] — there are always 3 groups of (n-1).
- The rest [green] makes a rectangle if I flip them and put them together; the dimensions of the rectangle are (n-2) and (n-1).
- My equation is T = 4(n-1) + 4 + 3(n-1) + (n-2)(n-1).
Student 2
I love how this kid saw this!!
- I see a square, or the toothpicks make up a square of (n+1) sides.
- Then there are always 2 left over groups of (n-1), so T = [(n+1)^2] + 2(n-1).
Student 3
- I see the perimeter as 4n.
- The inside diagonals [orange] are 3 groups of (n-1).
- The these two groups [purple] are Gauss, with (n-2) as the highest term.
- My equation is T = 4n + 3(n-1) + 2[1 + (n-2)][(n-2)/2].
Student 4
- I see this middle [green] rectangle with sides (n-1) and (n+2).
- Top and bottom are (n-1), then this extra [light blue] row is (n-1), plus the 4 corners.
- My equation is T = (n-1)(n+2) + 2(n-1) + (n-1) + 4.
pattern talk 18
Student 1
Student 2
Student 3
- I see n columns of 3. There are always 2 horizontal [green] groups of (n+1). Top and bottom [yellow] are each (n-1). The left over on the sides is always 6. My equation is T = 3n + 2(n+1) + 2(n-1) + 6.
Student 2
- I see the pattern growing from the middle out. So the outside has 5 toothpicks on each side, that’s always 10. Vertically, there are n groups of 3. Horizontally, there are 4 groups of (n-1). So, T = 10 + 3n + 4(n-1).
Student 3
- The middle here [red] are two groups of (n+1). Then these vertical pieces [green] are 3 groups of (n+2). Top and bottom make up the rest, each is (n-1). My equation is T = 2(n+1) + 3(n+2) + 2(n-1).
pattern talk 19
I tossed out this pattern for 2 reasons:
Student 1
Student 2
Student 3
- They’d already seen this pattern toward the beginning of our math talks — it was during week 3 — and I was curious to learn of students’ growth in pattern reasoning.
- Zack Patterson‘s students saw them as triangles [instead of toothpicks in the answer key] — and there were at least 3 ways to see the triangles — thus I was grateful he’d pointed this out.
Student 1
- I always saw 3 groups of consecutive numbers: along the horizontals and diagonals. For example, on step 3, I saw 3 groups of 1 + 2 + 3. My equation is T = 3(n + 1)(n/2).
Student 2
- I saw the base as always the same as step number, so base is n. Then I see the 2 sides also as n each. I did the table of values for the middle toothpicks. Step 1 had none, step 2 had 3, step 3 had 9, step 4 had 18, step 5 had 30 — but I ran out of time.
Student 3
- I just saw the pattern as stacks of triangles. So, step 4 would add 4 more triangles at the bottom. Step whatever would have that many triangles at the bottom, and on and on to the top with one triangle. Each triangle has 3 toothpicks, so it’s really Gauss equation times 3. My equation is T = 3([(n^2)+n]/2).
pattern talk 20
These are all from my 6th graders!! I’m so proud of them. This pattern is too easy apparently for my 8th graders.
Student 1
Student 2
Student 3
Student 4
Student 5
This one blew my socks off!
Student 1
- I always see this top one alone. Then I see 2 groups, one on each side, each group has twice the n number. But they share 1, so minus 1. My equation is D = 1 + 2(2n) – 1.
Student 2
- I see these 2 in the middle. The I see the outer wings as n each, the inner wings as (n – 1) each. My equation is D = 2 + 2n + 2(n – 1).
Student 3
- I have D = 2(n + 1) + 2(n – 1).
Student 4
- I see D = 4n. Hehe!
Student 5
This one blew my socks off!
- I see D = 4n too, but I rearranged the dots into a parallelogram: its base is always 4, its height always n.