Pattern talk 5
Student 1
Student 2
Student 3
Student 4
- I don’t know the equation but I think I can build it.
- For step 43, you’d build 43 triangles on the bottom layer, then 42 on top of that, then so on and so forth until 1 at the top.
Student 2
- I know the perimeter is always 3 times the step number. But I don’t know how to find the toothpicks inside the perimeter.
Student 3
- I used a table of values and got (n + 1)(n/2) for the number of triangles, then multiply this by 3 because each triangle has 3 toothpicks.
Student 4
- Oh, isn't what Student 3 shared the Gauss equation?
pattern talk 6
Student 1
Student 2
Student 3
Mrs. Nguyen
I haven’t shared with kids how I see any of the patterns, but we had some extra time so I shared with them my way.
- I see the top row as always (n + 1). The middle row is always (2n + 1). Then the next 2 rows are the same, each one is (2n – 1). So my equation is S = (n + 1) + (2n + 1) + 2(2n – 1).
Student 2
- I always see these 4 squares on the sides. Then the top middle is always (n – 1). Then I see columns of 3 tall: an (n) column and an (n – 1) column. My equation is S = 4 + (n – 1) + 3n + 3(n – 1).
Student 3
- I see the top row as (n + 1). Then there are always 2 leftover squares on the sides. The rest is a rectangle, always 3 high and (2n – 1) wide. My equation is S = (n + 1) + 2 + 3(2n – 1).
Mrs. Nguyen
I haven’t shared with kids how I see any of the patterns, but we had some extra time so I shared with them my way.
- I see each step as a full rectangle, with pieces missing. The full rectangle is always 4 tall and (2n + 1) wide. But there are always 4 missing squares on both lower sides. And the top row is always missing (n) squares. My equation is S = 4(2n + 1) – 4 – n.
pattern talk 7
Student 1
Student 2
Student 3
- I see a square of (n) sides, then a piece missing in the corner. The equation is S = (n^2) – 1.
Student 2
- I see a square of (n – 1) sides, then 2 groups of (n – 1) left over. My equation is S = [(n - 1)^2] + 2(n – 1).
Student 3
- I see the top row as always (n – 1). The remaining pieces form a rectangle of height (n – 1) and width (n). The equation is S = (n – 1) + n(n -1).
pattern talk 8
Student 1
Student 2
Student 3
- I see that it’s always (n) tall and (2n + 1) wide. So my equation is H = n(2n + 1).
Student 2
- I saw 2 sets of rectangles. One is always (n) tall and 2 wide — seen here in green, the other is always (n) tall and (2n – 1) wide. My equation is H = 2n + n(2n – 1).
Student 3
- I see 2 squares of (n by n). Then there’s (n) leftover, H = 2n^2 + n.