PT 21-24

pattern talk 21

Pattern #124

Student 1

I see the diagonal, then 3 groups of n on the outside, then the bottom group. My equation is D = (n+2) + 3n + (n+1).

Student 2

Left and right, each is n. Top and bottom, each is (n+2). The leftover dot is 1 less than n. My equation: D = 2n + 2(n+2) + (n-1).

Student 3

I saw these 4 in the right corner in every step. My equation is D = 4 + 2n + 2(n-1) + (n+1).

Pattern #126

Student 1

I see top and bottom as 2 groups of (n-1). The middle has (n+1) by (n). My equation is R = 2(n-1) + n(n+1).

Student 2

I move one rod from the bottom row to the top to make it a complete “rectangle” of (n+2) by (n). The leftover rods on the bottom row is (n-2). My equation is R = n(n+2) + (n-2).

Pattern #130

Student 1

I always see 2 groups of (n+1). The leftover dots are (n-1). My equation is D = 2(n+1) + (n-1).

Student 2

I see the bottom dot separately. The rest are n groups of 3. So, it’s D = 3n + 1.

Student 3

I saw step 1 in every step, so a constant of 4. Then I see (n-1) groups of 3. My equation is D = 4 + 3(n-1).

Student 4

I saw (n+1) groups of 3 dots. But there are always 2 [red] dots missing. So, D = 3(n+1) – 2.

Student 5

I see 4 groups of n. But there’ll be overlaps to subtract. That overlap is (n-1). My equation is D = 4n – (n-1).

Student 6

I always see two groups of 2 on the outside. The middle dots are (n-1) groups of 3. My equation is D = 2(2) + 3(n-1).

Pattern #106

Didn't realize we repeated a pattern. We did this already in Pattern Talk 15.

Student 1

I see a square of side n. The leftover is 2 groups of n with overlap of 1. My equation is A = n^2 + 2n – 1.

Student 2

I see 2 overlapping squares. The overlapped region is also a square. My equation is A = 2n^2 – (n-1)^2.

Student 3

Student 4

I see 2 groups of n on top and bottom. The middle is a rectangle of dimensions (n-1) and (n+1). My equation is A = 2n + (n-1)(n+1).