Consecutive Sums - Education Development · PDF file Consecutive Sums Student Dialogue...
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© 2016 by Education Development Center. Consecutive Sums is licensed under the Creative Commons Attribution- NonCommercial-NoDerivatives 4.0 International License. To view a copy of this license, visit https://creativecommons.org/licenses/by-nc-nd/4.0/. To contact the copyright holder email [email protected] This material is based on work supported by the National Science Foundation under Grant No. DRL-1119163. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
About Illustrations: Illustrations of the Standards for Mathematical Practice (SMP) consist of several pieces, including a mathematics task, student dialogue, mathematical overview, teacher reflection questions, and student materials. While the primary use of Illustrations is for teacher learning about the SMP, some components may be used in the classroom with students. These include the mathematics task, student dialogue, and student materials. For additional Illustrations or to learn about a professional development curriculum centered around the use of Illustrations, please visit mathpractices.edc.org. About the Consecutive Sums Illustration: This Illustration’s student dialogue shows the conversation among three students who are asked to think about how many different ways a number can be written using consecutive sums. As they explore different consecutive sums students are able to rewrite sums as products and find that any multiple of an odd number can be written as a consecutive sum. Highlighted Standard(s) for Mathematical Practice (MP) MP 1: Make sense of problems and persevere in solving them. MP 7: Look for and make use of structure. MP 8: Look for and express regularity in repeated reasoning. Target Grade Level: Grades 6–7 Target Content Domain: The Number System, Expressions & Equations Highlighted Standard(s) for Mathematical Content 6.NS.C.6a Recognize opposite signs of numbers as indicating locations on opposite sides of
0 on the number line; recognize that the opposite of the opposite of a number is the number itself, e.g., -(-3) = 3, and that 0 is its own opposite.
6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3(2+x) to produce the equivalent expression 6 +3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6(4x + 3y); apply properties of operations to y +y +y to produce the equivalent expression 3y.
7.NS.A.1b Understand p + q as the number located a distance |q| from p, in the positive or negative direction depending on whether q is positive or negative. Show that a number and its opposite have a sum of 0 (are additive inverses). Interpret sums of rational numbers by describing real-world contexts.
7.EE.A.2 Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. For example, a + 0.05a = 1.05a means that “increase by 5%” is the same as “multiply by 1.05.”
Math Topic Keywords: consecutive numbers, sums, factors
Mathematics Task Suggested Use This mathematics task is intended to encourage the use of mathematical practices. Keep track of ideas, strategies, and questions that you pursue as you work on the task. Also reflect on the mathematical practices you used when working on this task. The number 13 can be expressed as a sum of consecutive positive integers 6 + 7. Fourteen can be expressed as 2 + 3 + 4 + 5, also a sum of consecutive positive integers. Some numbers can be expressed as a sum of consecutive positive integers in more than one way. For example, 25 is 12 + 13 and is also 3 + 4 + 5 + 6 + 7. Look for evidence that would be useful in answering the questions: Is there a general rule for how many different ways a number can be written as a sum of consecutive positive integers? If there is a general rule, what is it? Task Source: Problem adapted from Benson, S., Addington, S., Arshavsky, N., Cuoco, A., Goldenberg, E. P., & Karnowski, E. (2005). Ways to think about mathematics. Thousand Oaks: Corwin Press.
Student Dialogue Suggested Use The dialogue shows one way that students might engage in the mathematical practices as they work on the mathematics task from this Illustration. Read the student dialogue and identify the ideas, strategies, and questions that the students pursue as they work on the task. This conversation occurs during the second day of working on this problem. Students have already made some discoveries about the problem and shared some insights during the previous day’s class. In particular, one group noticed that the sum of any three consecutive integers is equivalent to three times the middle number. Students begin with this idea on the second day. (1) Dana: We know a way to find the sum of any three consecutive numbers. We know 7 +
8 + 9 without even adding. We can just multiply: 8 times 3 is 24. (2) Sam: I think it works because the 7 and 9 sort of balance each other out—one of them is
1 less than 8 and the other is 1 more than 8. (3) Anita: So, let’s write it like this: [writes] (8 – 1) + 8 + (8 + 1). (4) Sam: Yeah! That’s what I said, but I like how you wrote it, Anita. You can see that the
negative and positive 1s cancel out so you’re left with 8 + 8 + 8. (5) Dana: I like it! What if we take it further?
(8 – 2) + (8 – 1) + 8 + (8 + 1) + (8 + 2) The negatives and positives cancel out again, so 6 + 7 + 8 + 9 + 10 must be 8 times 5, which is 40.
(6) Sam: And one step further would be
(8 – 3) + (8 – 2) + (8 – 1) + 8 + (8 + 1) + (8 + 2) + (8 + 3) That’s 5 + 6 + 7 + 8 + 9 + 10 + 11, or 8 times 7, which is 56. So, we’ve made 8 • 3, 8 • 5, and 8 • 7.
(7) Anita: But there’s nothing special about 8. Can we try this with another number like 11? (8) Dana: Sure. 11 • 3 = 33 and we can make 33 with (11 – 1) + 11 + (11 + 1), which is
10 + 11 + 12. Yup, that checks. (9) Anita: And 11 • 5 = 55, and that’s… well, I don’t want to write it out. But we need 5
numbers, 2 of them lower than 11 and 2 higher, so it starts with 11 – 2 and ends with 11 + 2. So we’re adding the numbers between 9 and 13: that’s 9 + 10 + 11 + 12 + 13. That’s… 55.
(10) Sam: And 11 • 7 = 77 and that sum starts with 11 – 3 and ends with 11 + 3. So that’s
8 + 9 + 10 + 11 + 12 + 13 + 14. I’m going to assume that’s 77. I mean it has to be because it’s the same as adding seven 11s together.
(11) Dana: Great, so now we can say that the sum of three consecutive numbers is 3 times the middle number. The sum of five consecutive numbers is 5 times the middle number. And the sum of 7 consecutive numbers is 7 times the middle number.
(12) Anita: That’s true, Dana. But instead of saying it that way, what if we look at it this way:
we just showed a way to make any multiple of 3, any multiple of 5, and any multiple of 7. I think we can even say that we can make any multiple of any odd number greater than 1.
(13) Sam: Whoa! I need to think about that for a minute.
[Students pause and think on their own.]
So if we can make any multiple of any odd number, can we also make any multiple of any even number?
(14) Anita: Wait… Maybe we can think about that in a minute, Sam. I think I’m onto
something else: What if a number is the multiple of 2 odd numbers? Take 15: it’s a multiple of 3 and a multiple of 5.
Since we can make any multiple of 3 with 3 numbers and any multiple of 5 with 5 numbers, that means we have two different ways of making 15:
15 = 5 + 5 + 5 = (5 – 1) + 5 + (5 + 1) = 4 + 5 + 6 and 15 = 3 + 3 + 3 + 3 + 3 = (3 – 2) + (3 – 1) + 3 + (3 + 1) + (3 + 2) = 1 + 2 + 3 + 4 + 5
(15) Dana: Yeah, that makes sense. Let me try it. I’ll use 33 because we already found one of
the sums: 10 + 11 + 12. So now if we think of this as a multiple of 11, then 3 is the middle number. Since we need 11 numbers, the biggest number is 3 + 5 and the smallest number is 3 – 5… Wait.
(16) Sam: We can’t use negative numbers. (17) Anita: Sam’s right. We’re only supposed to use positive number integers in this problem.
But there’s nothing wrong with the math. It should still work out… Try it, Dana. (18) Dana: Ok. So we have: (3 – 5) + (3 – 4) + (3 – 3) + (3 – 2) + (3 – 1) + 3 + (3 + 1) +
(3 + 2) + (3 + 3) + (3 + 4) + (3 + 5). That’s 11 numbers. It has to be 3 • 11 because we know all the positives and negatives cancel out.
(19) Sam: That means 33 is –2 + –1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. So I guess we just
extended the problems to include negative numbers. I’ll check it. –2 plus –1 is –3…
(20) Anita: Wait, there’s a shortcut. The positives and negatives cancel again! The first five
numbers add up to 0, so 33 is equal to the sum of the last six numbers: 3 + 4 + 5 + 6 + 7 + 8. Right?
(21) Sam: Wow. That worked! How else can we write 33? What about 1 • 33? Um… Never mind. That’s too many numbers.
(22) Dana: Wait. Let’s try it. I bet we can find some shortcuts. (23) Sam: Ok. So 1 • 33 is 1 + 1 + 1 + …