# Lesson 8

The $n^{\text{th}}$ Term

## 8.1: Which One Doesn’t Belong: Repeated Operations (5 minutes)

### Warm-up

This warm-up prompts students to compare four expressions. It gives students a reason to use language precisely (MP6). It gives the teacher an opportunity to hear how students use terminology and talk about characteristics of the items in comparison to one another.

The purpose of this activity is to remind students that repeated addition can be represented with multiplication and repeated multiplication can be represented with an exponent, which will help them make sense of some of the different ways equations for sequences can be written.

### Launch

Arrange students in groups of 2–4. Display the expressions for all to see. Give students 1 minute of quiet think time and then time to share their thinking with their small group. In their small groups, ask each student to share their reasoning why a particular item does not belong.

### Student Facing

Which one doesn’t belong?

A. \(5+2+2+2+2+2+2\)

B. \(5+6 \boldcdot 2\)

C. \(5 \boldcdot 2^6\)

D. \(5 \boldcdot 2\boldcdot 2\boldcdot 2\boldcdot 2\boldcdot 2\boldcdot 2\)

### Student Response

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### Activity Synthesis

Ask each group to share one reason why a particular item does not belong. Record and display the responses for all to see. After each response, ask the class if they agree or disagree. Since there is no single correct answer to the question of which one does not belong, attend to students’ explanations and ensure the reasons given are correct.

If it doesn’t come up during the discussion, highlight the fact that A and B are equivalent expressions, and C and D are equivalent expressions.

## 8.2: More Paper Slicing (15 minutes)

### Activity

In this activity students return to the 8 inch by 10 inch grid from an earlier lesson. This time, the goal is to work with non-recursive definitions of two different sequences, one geometric and one arithmetic, based on the grid being cut up in different ways. Students express regularity in repeated reasoning (MP8) by using their understanding of how the values of specific terms are calculated before explaining or expressing how the \(n^{\text{th}}\) term is calculated. This activity is meant as an introduction to the idea that domains of functions depend on how we think about the relationship and students will have several more opportunities to think critically about the starting term of a sequence.

Making graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).

### Launch

Arrange students in groups of 2. If using, distribute scissors and 2 copies of the blackline master to each group.

Ask students to recall the paper cutting activity from an earlier lesson. In it, Clare takes a sheet of paper that is 8 inches by 10 inches, cuts the paper in half, stacks the pieces, cuts the pieces in half, then stacks them, etc. We can let \(C(n)\) be the area, in square inches, of each piece based on the number of cuts \(n\). Display the table for all to see.

\(n\) (number of cuts) |
\(C(n)\) (area in square inches of each piece) |
---|---|

0 | 80 |

1 | 40 |

2 | 20 |

3 | 10 |

4 | 5 |

Depending on time, either invite students to share their observations about \(C(n)\) or tell students that it is a geometric series with growth factor \(\frac12\). Say, “This sequence starts with \(n=0\) since we start with a piece of paper with 0 cuts. How can we write a recursive definition for \(C(n)\)?” (\(C(0)=80, \)\(C(n)=C(n-1)\boldcdot \frac12,\)\( n\ge1\).) Select students to share their definitions paying particular attention to the starting term, \(C(0)\), and that \(n\ge1\) is used. An important takeaway from looking at this recursive definition is that the domain of a sequence is something that should be based on the situation and that sometimes starting with \(n=1\) doesn't make sense. Luckily, we can use the function notation to make clear that we are starting with \(n=0\) by beginning with \(C(0)\) and then we change \(n\ge2\) to \(n\ge1\) to match.

### Student Facing

- Clare takes a piece of paper with length 8 inches and width 10 inches and cuts it in half. Then she cuts it in half again, and again. . .
- Instead of writing a recursive definition, Clare writes \(C(n) = 80 \boldcdot \left(\frac 1 2\right)^n\), where \(C\) is the area, in square inches, of the paper after \(n\) cuts. Explain where the different terms in her expression came from.
- Approximately what is the area of the paper after 10 cuts?

- Kiran takes a piece of paper with length 8 inches and width 10 inches and cuts away one inch of the width. Then he does it again, and again. . .
- Complete the table for the area of Kiran’s paper \(K(n)\), in square inches, after \(n\) cuts.
\(n\) \(K(n)\) 0 80 1 2 \(80-8-8=80-8(2)=64\) 3 4 5 - Kiran says the area after 6 cuts, in square inches, is \(80 - 8 \boldcdot 6\). Explain where the different terms in his expression came from.
- Write a definition for \(K(n)\) that is not recursive.

- Complete the table for the area of Kiran’s paper \(K(n)\), in square inches, after \(n\) cuts.
- Which is larger, \(K(6)\) or \(C(6)\)?

### Student Response

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### Anticipated Misconceptions

Students who have trouble visualizing what's happening to the paper in each sequence may benefit from drawing the paper at each step and labeling it with dimensions, or cutting paper themselves and calculating the areas. In particular, if students don't see why Kiran removes 8 square inches each time, encourage them to write down the dimensions of the paper for the first few steps and calculate each area (and draw the paper at each step if needed).

### Activity Synthesis

Display the two definitions from this task for all to see: \(C(n)=80 \boldcdot \left(\frac12\right) ^n\) and \(K(n)=80 - 8n\). Begin the discussion by asking “How can you tell which of these defines a geometric sequence and which defines an arithmetic sequence?” (A geometric sequence has a constant growth factor between terms, so \(C(n)\) must represent a geometric sequence since each term is half the value of the previous term. An arithmetic sequence has a constant rate of change, so \(K(n)\) must represent an arithmetic sequence since each term is 8 less than the previous term.)

Tell students that \(C(n)\) and \(K(n)\) are examples of defining as sequence by the \(n^{\text{th}}\) term, which is the type of deinition students are likely most familiar with. These are sometimes known as a *closed-form* or *explicit* definition, but students do not need to use these terms. In future activities, if students are asked to represent a sequence with an equation for the \(n^{\text{th}}\) term, then they are being asked for this type of equation and not to define the sequence recursively. In later lessons, students will have practice writing equations for linear and exponential from a variety of situations.

Lastly, if time allows ask students to calculate which is larger, \(K(10)\) or \(C(10)\)? (\(C(10)\) is larger, since \(K(10) = 0\) and \(C(10) > 0\), or there can’t be a comparison because \(K(10)\) does not exist since a tenth cut is not possible.) Select students to share their calculations. If no student points out that \(K(10)\) does not exist due to the constraints of the context when \(n\) is the number of cuts, make sure to bring this point up. Students will have more opportunities to think directly about domains given a specific situation in a later lesson.

*Engagement: Develop Effort and Persistence.*Break the class into small group discussion groups and then invite a representative from each group to report back to the whole class.

*Supports accessibility for: Language; Social-emotional skills; Attention*

## 8.3: A Sierpinski Triangle (15 minutes)

### Activity

The goal of this activity is for students to understand that the way an equation is written to define a function depends on how the domain of the function is interpreted. In the previous activity, starting the sequence \(C\) with \(C(0)\) made sense since \(n\) was defined as the number of cuts while \(C(n)\) is the area of a paper square after the cuts. In this activity, students consider two equations written for a sequence represented by a visual pattern. One of the equations assumes the sequence starts at Step 0 and the other assumes the sequence starts at Step 1.

To help all students understand that both equations generate the same sequence and correctly represent the visual pattern, during the activity and whole-class discussion students should be encouraged to use precise language as they explain why an equation is valid (MP6). Confusion is likely to arise unless students are clear about the definition for \(n\) each equation is assuming. Some students may find the idea that both equations are equally valid challenging, so it is important to take time during the discussion for these students to understand that part of mathematics is recognizing what assumptions you are making and addressing them appropriately.

Monitor for students with clear definitions to select to share during the whole-class discussion. In particular, students that create different representations, such as tables, to see that the two equations result in the same list of numbers as their recursive definitions should be highlighted.

Making a spreadsheet available gives students an opportunity to choose appropriate tools strategically (MP5).

### Launch

Arrange students in groups of 2.

### Student Facing

A Sierpinski triangle can be created by starting with an equilateral triangle, breaking the triangle into 4 congruent equilateral triangles, and then removing the middle triangle. Starting from a single black equilateral triangle:

- Let \(S\) be the number of black triangles in Step \(n\). Define \(S(n)\) recursively.
- Andre and Lin are asked to write an equation for \(S\) that isn't recursive. Andre writes \(S(n)=3^n\) for \(n\ge0\) while Lin writes \(S(n)=3^{n-1}\) for \(n\ge 1\). Whose equation do you think is correct? Explain or show your reasoning.

### Student Response

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### Student Facing

#### Are you ready for more?

Here is a geometric sequence. Find the missing terms.

3, ____, 6, ____, 12, ____, 24

### Student Response

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### Anticipated Misconceptions

Students may assume that at least one person has to be wrong because their equations aren't equivalent. Encourage them to make a table for each equation that starts with the first value for \(n\) according to the definitions.

### Activity Synthesis

The goal of this discussion is for students to understand why Andre and Lin's equations are both representations of the visual pattern due to the interpretation of \(n\). Display the visual pattern for all to see. Invite previously selected students to share their reasoning about why either Andre or Lin is correct, recording student reasoning for all to see. If students still think starting with 0 is unusual after the share out, give some more time for students to reason about Andre's point of view, and then select students to share. (For example, Andre could be thinking about \(n\) as the breaking apart phase, so for the solid black triangle that has happened 0 times.)

Conclude the discussion by telling students that identifying an appropriate domain for a function is partly dependent on the situation and partly dependent on how they see the relationship. There are often many correct equations that represent a function. An important takeaway for students is that when they write an equation to represent a situation, they need to be clear what domain they have identified so other people can correctly interpret what they've done.

*Writing, Conversing: MLR1 Stronger and Clearer Each Time.*Use this routine to prepare students for the whole-class discussion by providing them with multiple opportunities to clarify their explanations through conversation. Before the whole-class discussion begins, give students time to meet with 1–2 other groups to share their response to the question, “Whose equation do you think is correct?” Invite listeners to ask questions, and to press for details and mathematical language.

*Design Principle(s): Optimize output (for explanation); Cultivate conversation*

## Lesson Synthesis

### Lesson Synthesis

Display these two sequences from an earlier lesson for all to see:

\(C\): 1, 2, 4, 8, 16, . . .

\(E\): 20, 13, 6, -1, -8, . . .

Invite students to pick one of the sequences and write an equation for the \(n^{\text{th}}\) term. If students are unsure where to start, remind them of their work writing equations for linear and exponential functions done in previous courses. Arithmetic and geometric sequences are just special types of these functions with a restricted domain. After some work time, select students to share their equations, making sure they clearly state whether they assumed the sequence started with \(n=0\) or \(n=1\). For example, an equation for \(E\) could be \(E(n)=20-7(n)\) for \(n\ge0\) or it could be \(E(n)=20-7(n-1)\) for \(n\ge1\).

## 8.4: Cool-down - Different Types of Equations (5 minutes)

### Cool-Down

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## Student Lesson Summary

### Student Facing

Here’s an arithmetic sequence \(f\): 6, 10, 14, 18, 22, . . . . In this sequence, each term is 4 more than the previous term. One recursive definition of this sequence is \(f(1)=6\), for \(f(n)=f(n-1)+4\) for \(n\ge2\). We could also write \(f(0)=6,\) \(f(n)=f(n-1)+4,\) for \(n\ge1\) since it generates the same sequence. Neither of these definitions is better than the other, we just have to remember how we chose to define the “first term” of the sequence: \(f(1)\) or \(f(0)\). Let's use \(f(1)\) for now.

While defining a sequence recursively works to calculate the current term from the previous, if we wanted to calculate, say, \(f(100)\), it would mean calculating all the terms up to \(f(99)\) to get there! Let's think of a better way.

Since we know that each term has an increasing number of fours, we could write the terms of \(f\) organized in a table like the one shown here.

\(n\) | \(f(n)\) |
---|---|

1 | \(6+0=6+4(0)=6\) |

2 | \(6+4=6+4(1)=10\) |

3 | \(6+4+4=6+4(2)=14\) |

4 | \(6+4+4+4=6+4(3)=18\) |

5 | \(6+4+4+4+4=6+4(4)=22\) |

Looking carefully at the pattern in the table, we can say that for the \(n^{\text{th}}\) term \(f(n)=6+4(n-1)\) for \(n\ge1.\) This is sometimes called an *explicit* or *closed-form definition *of a sequence, but it's really just a way to calculate the value of the \(n^{\text{th}}\) term without having to calculate all the terms that came before it. Need to know \(f(100)\)? Just compute \(6+4(100-1).\) Defining an arithmetic sequence this way takes advantage of the fact that this type of sequence is a linear function with a starting value (in this case 6) and rate of change (in this case 4). If we had decided to start the sequence at \(n=0\) so that \(f(0)=6,\) we would have written the equation for the \(n^{\text{th}}\) term as \(f(n)=6+4(n)\) for \(n\ge0.\)

Geometric sequences behave the same way, but with repeated multiplication. The geometric sequence \(g\): 3, 15, 75, 375, . . . can be written as \(3,3\boldcdot5,3\boldcdot5\boldcdot5, 3\boldcdot5\boldcdot5\boldcdot5, . . .\) This means if \(g(0)=3\), we can define the \(n^{\text{th}}\) term directly as \(g(n)=3\boldcdot5^n.\)