PT 25-28

pattern talk 25

Pattern #44

Student 1

I see the shape as 2 halves, top and bottom. For each half, I see n groups of 3. But the groups overlap, and the overlap is always (n-1). My equation is H = 2(3n) – 2(n-1).

Student 2

I see the two middle rows of hexagons of (n+1) each. Then there are leftover groups of n on top and bottom: H = 2(n+1) + 2n.

Student 3

I always see 2 hexagons at the front. Then I see n groups of 5, but that includes the white inside hexagons, so I have to subtract them — and there are always n white ones. My equation is H = 2 + 5n – n.

Student 4

I see n groups of 6. But there are (n-1) overlaps of 2 each. My equation is H = 6n – 2(n-1).

Pattern #52

Student 1

I moved the front pieces to the left side and the back pieces to the right side, so it’s one flat rectangle. The equation is C = (n+1)(2n+1).

Student 2

I see the center column as (n+1). Then there are 4 identical groups around the center, each one is a Gauss addition. My equation is C = (n+1) + 4[(n2+n)/2].

Student 3

I see the center column as (n+1). Then I rearranged the front and back pieces together to form a rectangle. I do the same with the left and right pieces. My equation is C = (n+1) + n(n+1) + n(n+1).

Similar to how Student 1 sees it, but she creates 2 separate rectangles instead of 1.

Student 4

I take the left and right pieces, including the middle column and flatten them out into a rectangle like everyone else has done, but I see that I can move the pieces to form a square of (n+1) sides. So I have 2 of these because I do the same to the front and back pieces. Because I count the middle column of (n+1) for both squares, so I have to subtract it from my equation: C = 2(n+1)2 – (n+1).

Pattern #136

Student 1

Student 2

I see a full rectangle, then subtract the empty space: S = 2n(2n+1) – n(n+1).

Student 3

Student 4

I move the whole left portion down to bottom of the right portion to have 1 rectangle: S = n(3n+1).

Pattern #134

Student 1

I see the middle rectangle separately from the 2 sides.
Then I move the left side pieces to join the right side pieces to form 1 rectangle: S = n + (n^2) + n(n+1).

Student 2

I move the left pieces over to the right to form one large rectangle: S = 2n(n+1).

Student 3

I see a Gauss equation. I add the top and bottom row, which is (n+3n). The number of pairs of with this 4n sum is the height of the pattern divided by 2: S = 4n[(n+1)/2].