Today we looked at Pattern #5 from visualpatterns.org.

I was intrigued by this one because I thought they looked like crab claws. Anyway, what was fascinating about this one was that the kids did not immediately expect it to be quadratic.

They came up with the following pre-simplified expressions for the nth step:

- 2n(n+1) + 3
- 1 + (n^2 +n^2) + (n+1) + (n+1)
- 2[n(n+1)] + 3
- 2n(n+1) + 3
- 3 + [(n+1)n] + [(n+1)n]
- 3+2(n+1) + 2[(n+1)(n-1)] **This one was the most intriguing to me.
- 2n^2 + 2n + 3

For each of these, I had the student put the expression on the board. I then had different students explain the thinking of the student who came up with the expression and relate it to the pictured pattern. I saw a real improvement here from when I had them do this activity the first time last week. I had many more students volunteer to explain the thinking of their cohorts and much less hesitation to work out what the terms in the expressions represented. The students sort of thought of ‘explaining thinking’ that was only represented by numbers and n’s was like decoding a puzzle. They could see that they all simplified into the same final expression, but working backward to find where that expression started was part of the challenge that they have been more willing to accept.

Here’s a picture of some of their work. I’m sure they’d be thrilled to be part of my blog post.

What was REALLY special about this pattern is that we were able to relate it then to quadratic equations in x,y coordinates. We talked about all kinds of things related to quadratics such as what the +3 means in the pattern and in the quadratic graph, how we could use the “0” step to make finding the quadratic equation easier, and how a system of equations could be set up as well to find the coefficients of the x^2 and x term.

What I really appreciate about this website is that there are so many extensions to all of the patterns. There is no much more for students to uncover other than finding the 43rd term. I love that I am able to use these patterns in multiple levels of my math classes and the students are given the opportunity to pull out the necessary mathematics.

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I think the fact that the students were “more comfortable” this time is a testament to the value of this style of learning. Your class sounds like a really good learning place!

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Your notes show lots of attempts at an explicit formula for the pattern, but its nature *might* be revealed a little quicker if you tried a recursive definition first.

You gave , , and which gives common differences of 8 and 12. Each common difference is 4 times the term number, so and , or generally, $t_n=t_{n-1}+4n$. Whenever the common difference is linear, the explicit formula must be quadratic. All that remains is to work out the coefficients.

Great series of problems. Thanks for sharing.

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