Pattern talk 13
Pattern #70
The equation for the blue parallelograms was straightforward enough as B = n – 1, so we focused our attention on discussing just the orange squares.
No one came up with the equation for whatever pattern they saw for the orange squares. Most common thinking was was 4, 1, 0, 1, 4, 1, 0, 1, …. One student thought it might be 4, 1, 0, 1, 4, 2, 0, 2, 4, 3, 0, 3, ….; others felt it was okay to see it that way too.
In the most common pattern seen, they figured that in every even-numbered step, the number of orange squares will always be 1. To figure out step 43, you’d take 43 divide by 4 because it the pattern repeats after every 4, leaving a remainder of 3 and that puts us at 0 orange squares.
Because no one came up with an equation, I told them I had an equation that would work and that I saw 9 orange squares in step 6. Not enough information for them yet, so I gave them 16 orange squares for step 7, and a few guessed that the next steps must be 25, 36, 49, 64, etc. Then someone said, it’s n minus 3, then square that.
No one came up with the equation for whatever pattern they saw for the orange squares. Most common thinking was was 4, 1, 0, 1, 4, 1, 0, 1, …. One student thought it might be 4, 1, 0, 1, 4, 2, 0, 2, 4, 3, 0, 3, ….; others felt it was okay to see it that way too.
In the most common pattern seen, they figured that in every even-numbered step, the number of orange squares will always be 1. To figure out step 43, you’d take 43 divide by 4 because it the pattern repeats after every 4, leaving a remainder of 3 and that puts us at 0 orange squares.
Because no one came up with an equation, I told them I had an equation that would work and that I saw 9 orange squares in step 6. Not enough information for them yet, so I gave them 16 orange squares for step 7, and a few guessed that the next steps must be 25, 36, 49, 64, etc. Then someone said, it’s n minus 3, then square that.
pattern talk 14
Pattern #28
Student 1
Student 2
Student 3
Student 4
- I see the floor and the 2 walls sharing a corner. My equation is C = (n-1)^2 + (2n^2) – n.
Student 2
- I see a full cube of n sides. But it’s not a full cube, so missing is a smaller cube of (n – 1) sides: C = (n^3) – (n-1)^3.
Student 3
- I see the entire floor as n^2, then the 2 walls, and the corner column. My equation is C = (n^2) + 2(n-1)^2 + (n-1).
Student 4
- I see the exposed floor part. Then I see the left and right walls as one piece, so C = (n-1)^2 + n(2n – 1).
pattern talk 15
Pattern #106
Student 1
Student 2
Student 3
Student 4
- I see a full square then 2 corner pieces missing. My equation is S = (n + 1)^2 – 2.
Student 2
- Top and bottom rows are n each. The middle rectangle is (n-1) by (n+1). My equation then is S = 2n + (n-1)(n+1).
Student 3
- Mine is similar to Student 2's, but I thought of the middle as a square with (n-1) sides and 2 groups of (n-1): S = 2n + (n-1)^2 + 2(n-1).
Student 4
- I see two overlapping squares of n sides. Then just subtract the overlap: S = 2n^2 – (n-1)^2.
pattern talk 16
Pattern #68
Student 1
Student 2
Student 3
- I see a full thing then 2 corner pieces missing. My equation is P = (n + 1)^2 – 2.
Student 2
- You really don’t see it, but what I did was I moved these 3 pieces on the bottom to the white space, and that forms a square. Then the right column is always n+1. My equation is P = (n^2) + (n+1).
Student 3
- I see (n+1) groups of n, and the corner square. My equation is P = n(n+1) + 1.
