## pattern talk 1

**Student 1**

- I see the perimeter first, so step n has 4n for perimeter.
- Then I see 1 fewer columns than the step number inside, each column has n toothpicks, so n(n-1).
- The rows are the same as the columns, so I just multiply what I have for columns by 2.
- My equation is:
**T = 4n + 2[n(n-1)]**.

**Student 2**

- I see the number of columns is always 1 more than the step number, and each column has the same number of toothpicks as step number, so I have n(n+1).
- This is true for the rows also, so I just multiply this quantity by 2 to get
**T = 2[n(n+1)]**.

**Student 3**

- I used an input/output table to find the common differences to get
**T = 2n^2 + 2n**.

## Pattern talk 2

**Student 1**

- I saw the number of blocks on the bottom is the same as step number.
- And I also see the same vertically, but they share the corner block, so it’s
**2n – 1**.

**Student 2**

- I saw n blocks horizontally, and n-1 blocks vertically on top. My equation is
**B = n + (n-1)**.

**Student 3**

- I ignored the corner block and saw that the horizontal blocks are always one less than the step number. Same thing vertically. Then I added the corner block. My equation is
**B = 2(n-1) + 1**.

## Pattern talk 3

**Student 1**

- I don’t have the answer to step 43 yet, but I have the equation as
**(n+1) + n + (n-1) + (n-1)**. - And this is how I got the equation: I tried the equation with the other steps, and it worked.

**Student 2**(

*in response to Student 1*)

- Really? Try it with step 3 right now.

**Student 3**

- I noticed that each row just goes up by one. I forgot what it’s called, and in talking with David, he reminded me of Gauss. So I have
**(n + 1)(n/2)**. I tried this for steps 1 through 5, and it worked.

*Her equation reflects “Gauss addition, but it’s not entirely correct for this pattern as the pattern is really “Gauss plus one addition,” or (n + 2)[(n + 1)/2].*

**Student 4**

- I drew a diagonal like this to get a triangle, and in step 4, there are 5 vertically and 5 horizontally. My equation for the triangle part is
**[(n + 1)^2]/2**. then the left over part is 5 halves, so I add**(n + 1)/2**.

**Student 5**

- I did an input/output table and knew it had to be squared, then I just came up with
**[(n^2) + 3n + 2]/2**.

## pattern talk 4

**Student 1**

- I see a column of n + 1, and another one up here. Then I see 2 squares, each side is always n. Plus this extra one tucked in here. So my equation is
**C = 2(n +1) + 2n^2 + 1**.

**Student 2**

- So I see on the left here n + 1. Then I see a square of n by n. Then I see a rectangle of n by (n + 1). And these 2 leftovers. My equation is
**C = n + 1 + n^2 + n(n + 1) + 2**.

**Student 3**

- I see two rectangles of the same size. Each one is n by (n + 1). And then the 3 circles here. My equation is
**C = 2[n(n + 1)] + 3**.