pattern talk 1

Pattern #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.
  

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Pattern talk 2

Pattern #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.

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Pattern talk 3

Pattern #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.

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pattern talk 4

Pattern #5

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.