## Engage NY Eureka Math Algebra 1 Module 3 Lesson 2 Answer Key

## Eureka Math Algebra 1 Module 3 Lesson 2 Example Answer Key

Example 1.

Consider Akelia’s sequence 5, 8, 11, 14, 17, ….

a. If you believed in patterns, what might you say is the next number in the sequence?

Answer:

20 (adding 3 each time)

b. Write a formula for Akelia’s sequence.

Answer:

A(n) = 5 + 3(n – 1)

c. Explain how each part of the formula relates to the sequence.

Answer:

To find each term in the sequence, you are adding 3 one less time than the term number. To get the 1st term, you add three zero times. To get the 2nd term, you add 3 one time. To get the 5th term, you add 3 four times.

d. Explain Johnny’s formula.

Answer:

His formula is saying that to find any term in the sequence, just add 3 to the term before it. For example, to find the 12th term, add 3 to the 11th term: A(12) = A(11) + 3. To find the 50th term, add 3 to the 49th term. To find the (n + 1)th term, add 3 to the nth term. It is critical that the value of the very first term be specified; we need it to get started finding the values of all the other terms.

Example 2.

Consider a sequence given by the formula a_{n} = a_{(n-1)}-5, where a_{1} = 12 and n ≥ 2.

a. List the first five terms of the sequence.

Answer:

12, 7, 2, -3, -8

b. Write an explicit formula.

Answer:

a_{n} = 12-5(n-1) for n ≥ 1

c. Find a_6 and a_100 of the sequence.

Answer:

a_{6} = -13 a_{100} = -483

### Eureka Math Algebra 1 Module 3 Lesson 2 Exercise Answer Key

Exercises 1–2

Exercise 1.

Akelia, in a playful mood, asked Johnny: “What would happen if we change the ‘ + ’ sign in your formula to a ‘-’ sign? To a ‘×’ sign? To a ‘ ÷ ’ sign?”

a. What sequence does A(n + 1) = A(n)-3 for n ≥ 1 and A(1) = 5 generate?

Answer:

Answer:

5, 2,-1, -4, …

b. What sequence does A(n + 1) = A(n) ⋅ 3 for n ≥ 1 and A(1) = 5 generate?

Answer:

5, 15, 45, 135, …

c. What sequence does A(n + 1) = A(n) ÷ 3 for n ≥ 1 and A(1) = 5 generate?

Answer:

5, \(\frac{5}{3}\), \(\frac{5}{9}\), \(\frac{5}{27}\), …

Exercise 2.

Ben made up a recursive formula and used it to generate a sequence. He used B(n) to stand for the nth term of his recursive sequence.

a. What does B(3) mean?

Answer:

It is the third term of Ben’s sequence.

b. What does B(m) mean?

Answer:

It is the m^{th} term of Ben’s sequence.

c. If B(n + 1) = 33 and B(n) = 28, write a possible recursive formula involving B(n + 1) and B(n) that would generate 28 and 33 in the sequence.

Answer:

B(n) = B(n-1) + 5 (Note that this is not the only possible answer; it assumes the sequence is arithmetic and is probably the most obvious response students will give. If the sequence were geometric, the answer could be written as B(n + 1) = (\(\frac{33}{28}\))B(n).)

d. What does 2B(7) + 6 mean?

Answer:

It is 2 times the 7th term of Ben’s sequence plus 6.

e. What does B(n) + B(m) mean?

Answer:

It is the sum of the nth term of Ben’s sequence plus the m^{th} term of Ben’s sequence.

f. Would it necessarily be the same as B(n + m)?

Answer:

No, adding two terms of a sequence is not the same as adding two of the term numbers and then finding that term of a sequence. Consider, for example, the sequence 1, 3, 5, 7, 9, 11, 13, …. Adding the 2nd and 3rd terms does not give you the 5th term.

g. What does B(17)-B(16) mean?

Answer:

It is the 17th term of Ben’s sequence minus the 16th term of Ben’s sequence.

Exercises 3–6

Exercise 3.

One of the most famous sequences is the Fibonacci sequence:

1, 1, 2, 3, 5, 8, 13, 21, 34, ….

f(n + 1) = f(n) + f(n – 1), where f(1) = 1, f(2) = 1, and n ≥ 2

How is each term of the sequence generated?

Answer:

By adding the two preceding terms

Exercise 4.

Each sequence below gives an explicit formula. Write the first five terms of each sequence. Then, write a recursive formula for the sequence.

a. a_{n} = 2n + 10 for n ≥ 1

Answer:

12, 14, 16, 18, 20

a_{n + 1} = a_{n} + 2, where a_{1} = 12 and n ≥ 1

b. a_n = (\(\frac{1}{2}\))^{(n-1)} for n ≥ 1

Answer:

1, \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\)

a_{n + 1} = a_{n} ÷ 2, where a_{1} = 1 and n ≥ 1

Exercise 5.

For each sequence, write either an explicit or a recursive formula.

a. 1, -1, 1, -1, 1, -1, …

Answer:

a_{(n + 1)}-a_{n}, where a_{1} = 1 and n≥1 or f(n) = (-1)^{(n + 1)}, where n ≥ 1

b. \(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{3}{4}\), \(\frac{4}{5}\), …

Answer:

f(n) = \(\frac{n}{n + 1}\) and n ≥ 1

Exercise 6.

Lou opens a bank account. The deal he makes with his mother is that if he doubles the amount that was in the account at the beginning of each month by the end of the month, she will add an additional $5 to the account at the end of the month.

a. Let A(n) represent the amount in the account at the beginning of the nth month. Assume that he does, in fact, double the amount every month. Write a recursive formula for the amount of money in his account at the beginning of the (n + 1)th month.

Answer:

A(n + 1) = 2A(n) + 5, where n ≥ 1 and A(1) is the initial amount

b. What is the least amount he could start with in order to have $300 by the beginning of the third month?

Answer:

A(3) = 2 ∙ A(2) + 5

A(3) = 2 ∙ [2 ∙ A(1) + 5] + 5

300 ≤ 2 ∙ [2 ∙ A(1) + 5] + 5

300 ≤ 4 ∙ A(1) + 15

71.25 ≤ A(1)

The least amount he could start with in order to have $300 by the beginning of the third month is $71.25.

### Eureka Math Algebra 1 Module 3 Lesson 2 Problem Set Answer Key

For Problems 1–4, list the first five terms of each sequence.

Question 1.

a_{n + 1} = a_{n} + 6, where a_{1} = 11 for n ≥ 1

Answer:

11, 17, 23, 29, 35

Question 2.

a_{n} = a_{n-1} ÷ 2, where a_{1} = 50 for n ≥ 2

Answer:

50, 25, 12.5, 6.25, 3.125

Question 3.

f(n + 1) = -2f(n) + 8 and f(1) = 1 for n ≥ 1

Answer:

1, 6, -4, 16, -24

Question 4.

f(n) = f(n-1) + n and f(1) = 4 for n ≥ 2

Answer:

4, 6, 9, 13, 18

For Problems 5–10, write a recursive formula for each sequence given or described below.

Question 5.

It follows a plus one pattern: 8, 9, 10, 11, 12,

Answer:

f(n + 1) = f(n) + 1, where f(1) = 8 and n ≥ 1

Question 6.

It follows a times 10 pattern: 4, 40, 400, 4000, ….

Answer:

f(n + 1) = 10f(n), where f(1) = 4 and n ≥ 1

Question 7.

It has an explicit formula of f(n) = -3n + 2 for n ≥ 1.

Answer:

(n + 1) = f(n)-3, where f(1) = -1 and n ≥ 1

Question 8.

It has an explicit formula of f(n) = -1(12)^{(n-1)} for n ≥ 1.

Answer:

f(n + 1) = 12f(n), where f(1) = -1 for n ≥ 1

Question 9.

Doug accepts a job where his starting salary is $30,000 per year, and each year he receives a raise of $3,000.

Answer:

D_{(n + 1)} = D_{n} + 3000, where D_{1} = 30000 and n ≥ 1

Question 10.

A bacteria culture has an initial population of 10 bacteria, and each hour the population triples in size.

Answer:

B_{(n + 1)} = 3B_{n}, where B_{1} = 10 and n ≥ 1

### Eureka Math Algebra 1 Module 3 Lesson 2 Exit Ticket Answer Key

Question 1.

Consider the sequence following a minus 8 pattern: 9, 1, -7, -15, ….

a. Write an explicit formula for the sequence.

Answer:

f(n) = 9-8(n-1) for n ≥ 1

b. Write a recursive formula for the sequence.

Answer:

f(n + 1) = f(n)-8 and f(1) = 9 for n ≥ 1

c. Find the 38th term of the sequence.

Answer:

f(38) = 9-8(37) = -287

Question 2.

Consider the sequence given by the formula a(n + 1) = 5a(n) and a(1) = 2 for n ≥ 1.

a. Explain what the formula means.

Answer:

The first term of the sequence is 2. Each subsequent term of the sequence is found by multiplying the previous term by 5.

b. List the first five terms of the sequence.

Answer:

2, 10, 50, 250, 1250