Precalculus by Richard Wright

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What good will it be for someone to gain the whole world, yet forfeit their soul? Or what can anyone give in exchange for their soul? Matthew 16:26 NIV

# 10-01 Sequences

Summary: In this section, you will:

• Write a sequence from a rule.
• Write an explicit rule for a sequence.
• Write a recursive rule for a sequence.
• Simplify factorial expressions.

SDA NAD Content Standards (2018): PC.7.2

The story goes like this. An ancient Indian minister named Sessa ibn Dahir invented the game of chess. His king liked the game and offered a reward. Sessa asked the king for a grain of wheat on the the first square of the board, 2 grains on the second square, 4 grains on the third square, and so on doubling the number of grains on each square. The king laughed at such a small reward. But was it a small request? Sequences and their explicit rules can be used to find out.

A sequence is a list of numbers that follows some sort of mathematical rule. If the sequence ends, then it is a finite sequence. Sequences that continue forever are called infinite.

Finite Sequence: 1, 2, 4, 8, 16

Infinite Sequence: 1, 2, 4, 8, 16, …

Sequences are part of discrete math where there are no fractional parts of the domain. Sequences are defined by mathematical rules such as an = 2n−1, where n is the domain (like x) and an is the range (like y). There are no fractional parts of n; there is 1 and 2, but no 1.5. The rule that gives the term value directly is called the explicit rule or rule for the nth term.

## Explicit Rules

###### Write a Sequence from an Explicit Rule
1. Plug in numbers for n, usually 1, 2, 3, 4, …
2. This gives a1, a2, a3, a4, …

#### Write a Sequence from an Explicit Rule

Write the first five terms of the sequence an = 3(−1)nn.

Solution

Let n = 1.

a1 = 3(−1)11

a1 = −4

Continue plugging in numbers for n.

a2 = 3(−1)22 = 1

a3 = 3(−1)33 = −6

a4 = 3(−1)44 = −1

a5 = 3(−1)55 = −8

The sequence is −4, 1, −6, −1, −8, …

Find the first five terms of an = 3n + 1.

4, 7, 10, 13, 16

###### Find the Explicit Rule

Given the first few terms of a sequence, find an explicit formula for the sequence using the following suggestions. These are not step-by-step instructions.

• Look for a pattern among the terms.
• If the terms are fractions, look for a separate pattern for the numerator and denominator.
• Look for a pattern among the signs of the terms. Alternating signs are created by (−1)n if a1 is negative or (−1)n−1 is a1 is positive.
• Write a formula for an in terms of n. Test your formula for n = 1, n = 2, and n = 3.

#### Find an Explicit Rule (Addition Pattern)

Find the rule for the nth term of 3, 5, 7, 9, 11, …

Solution

There is a pattern of adding 2.

 3, 5, 7, 9, 11, … 3, 3 + 2, 3 + 2 + 2, 3 + 2 + 2 + 2, 3 + 2 + 2 + 2 + 2, … 3 + 2·0, 3 + 2·1, 3 + 2·2, 3 + 2·3, 3 + 2·4, …

These expressions are the an. Compare them to the n's to find the rule.

 n an 1 2 3 4 5 3 + 2·0, 3 + 2·1, 3 + 2·2, 3 + 2·3, 3 + 2·4, …

Notice the multiples of 2 are 1 less than n. So, the rule for the nth term is

an = 3 + 2·(n − 1)

Distribute and simplify to get

an = 2n + 1

#### Find an Explicit Rule (Multiplication Pattern and Alternating Sign)

Find the rule for the nth term of 3, −6, 12, −24, 48, …

Solution

There is a pattern of multiplying by 2.

 3, −6, 12, −24, 48, … 3, −3·2, 3·2·2, −3·2·2·2, 3·2·2·2·2, … 3·20, −3·21, 3·22, −3·23, 3·24, …

These expressions are the an. Compare them to the n's to find the rule.

 n an 1 2 3 4 5 3·20 −3·21 3·22 −3·23 3·24

Notice the exponents are 1 less than n. There is also a pattern of alternating signs with a1 being positive so use (−1)n−1. So, the rule for the nth term is

an = 3·2n-1·(−1)n−1

#### Find an Explicit Rule (Fraction)

Find the rule for the nth term of $$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, …$$

Solution

When the terms are fractions, then treat the numerator and denominator separately. This pattern is adding 1 in the numerator and adding 1 in the denominator. However, these numbers are very close to n, so start by comparing the the an to the n.

 n an 1 2 3 4 5 $$\frac{1}{2}$$ $$\frac{2}{3}$$ $$\frac{3}{4}$$ $$\frac{4}{5}$$ $$\frac{5}{6}$$

Notice the numerators are equal to n. The denominators are 1 more than n. So, the rule for the nth term is

$$a_n = \frac{n}{n+1}$$

#### Find an Explicit Rule (Other)

Find the rule for the nth term of 0, 3, 8, 15, 24, …

Solution

There is no constant pattern or adding or multiplying to get the next number. Start by comparing the the an to the n.

 n an 1 2 3 4 5 0 3 8 15 24

Notice that the terms are close to perfect squares.

 n an an + 1 1 2 3 4 5 0 3 8 15 24 0 + 1 = 1 3 + 1 = 4 8 + 1 = 9 15 + 1 = 16 24 + 1 = 25

Notice the an + 1 = n2. So, the rule for the nth term is

an = n2 − 1

Find the rule for the nth term.

1. 3, 7, 11, 15, 19, …
2. $$\frac{2}{1}, \frac{4}{2}, \frac{6}{4}, \frac{8}{8}, \frac{10}{16}, …$$
3. 1, 8, 27, 64, 125, …

a. 4n − 1, b. $$\frac{2n}{2^{n-1}}$$, c. n3

## Recursive Rules

Explicit rules give the term, an, directly by simply plugging in the n. Recursive rules give the next term based on the previous term. Recursive rules are often easier to write, but they can be a nuisance to use. To find the 10th term, you have to find the 9th term. To get the 9th term, you have to find the 8th term. So, to find the 10th term, you must find all the terms starting from the 1st term.

For a recursive rule, an refers to the current term. an − 1 refers to the previous term.

#### Use a Recursive Rule

Write the first five terms of a1 = 4, an = 2an−1 − 1.

Solution

a1 is already given, so start by plugging in n = 2.

a1 = 4

an = 2an−1 − 1

a2 = 2a2−1 − 1

a2 = 2a1 − 1

a1 is 4.

a2 = 2(4) − 1 = 7

Plug in n = 3.

a3 = 2a3−1 − 1

a3 = 2a2 − 1

a3 = 2(7) − 1 = 13

Plug in n = 4.

a4 = 2a4−1 − 1

a4 = 2a3 − 1

a4 = 2(13) − 1 = 25

You can continue plugging in numbers, or the pattern can be used. The pattern for this sequence is 2 times the previous term − 1.

a5 = 2(25) − 1 = 49

The sequence is 4, 7, 13, 25, 49.

###### Write a Recursive Rule

an refers to the current term. an − 1 refers to the previous term.

1. Write an expression for the term based on the previous term such as
an = 2an−1 − 1
2. Write the first term so we know where to start such as
a1 = 4
3. Put those together
a1 = 4, an = 2an−1 − 1

#### Write a Recursive Rule

Write a recursive rule for 3, 5, 7, 9, 11, …

Solution

an = an−1 + 2

Don't forget the first term.

an = 3, an = an−1 + 2

Write a recursive rule for 2, 6, 18, 54, 162, …

a1 = 2, an = 3an−1

## Factorial

An important concept that appears in discrete math and probability is the factorial whose symbol is !. A factorial is a number multiplied by every natural number less than it. For example,

5! = 5 · 4 · 3 · 2 · 1 = 120

And by definition 0! = 1, which is important for probability.

###### Factorial

To find a factorial, multiply that number by every natural number less than it.

6! = 6 · 5 · 4 · 3 · 2 · 1 = 720

By definition 0! = 1

To simplify a factorial expression, write out enough terms to see a pattern, then simplify the pattern.

#### Simplify Factorial Expressions

Simplify $$\frac{10!n!}{2!8!(n+1)!}$$.

Solution

Begin by writing out each factorial.

$$\frac{\color{blue}{10!}\color{red}{n!}}{\color{purple}{2!}\color{green}{8!}\color{indigo}{(n+1)!}}$$

$$\frac{\color{blue}{10·9·8·7·6·5·4·3·2·1}·\color{red}{n(n-1)(n-2)\cdots}}{\color{purple}{2·1}·\color{green}{8·7·6·5·4·3·2·1}·\color{indigo}{(n+1)n(n-1)(n-2)\cdots}}$$

Cancel numbers.

$$\require{cancel} \frac{10·9·\cancel{8·7·6·5·4·3·2·1}·\cancel{n(n-1)(n-2)\cdots}}{2·1·\cancel{8·7·6·5·4·3·2·1}·(n+1)\cancel{n(n-1)(n-2)\cdots}}$$

$$\frac{90}{2(n+1)}$$

$$\frac{45}{n+1}$$

Simplify $$\frac{2!3!}{4!}$$

$$\frac{1}{2}$$

##### Lesson Summary

###### Write a Sequence from an Explicit Rule
1. Plug in numbers for n, usually 1, 2, 3, 4, …
2. This gives a1, a2, a3, a4, …

###### Find the Explicit Rule

Given the first few terms of a sequence, find an explicit formula for the sequence using the following suggestions. These are not step-by-step instructions.

• Look for a pattern among the terms.
• If the terms are fractions, look for a separate pattern for the numerator and denominator.
• Look for a pattern among the signs of the terms. Alternating signs are created by $$(-1)^{n}$$ if a1 is negative or $$(-1)^{n-1}$$ is a1 is positive.
• Write a formula for an in terms of n. Test your formula for n = 1, n = 2, and n = 3.

###### Write a Recursive Rule

an refers to the current term. an − 1 refers to the previous term.

1. Write an expression for the term based on the previous term such as
an = 2an−1 − 1
2. Write the first term so we know where to start such as
a1 = 4
3. Put those together
a1 = 4, an = 2an−1 − 1

###### Factorial

To find a factorial, multiply that number by every natural number less than it.

6! = 6 · 5 · 4 · 3 · 2 · 1 = 720

By definition 0! = 1

To simplify a factorial expression, write out enough terms to see a pattern, then simplify the pattern.

## Practice Exercises

1. Write the first 5 term of the sequence.
2. an = −2n + 1
3. an = n2 + n
4. $$a_n = (-1)^n \left(\frac{n}{n+2}\right)$$
5. a1 = 2, an = 2an−1 + 3
6. a1 = −3, an = (an−1)2
7. Write the rule for the nth term.
8. 1, 5, 9, 13, 17, …
9. $$\frac{1}{2}, \frac{3}{4}, \frac{5}{8}, \frac{7}{16}, \frac{9}{32}, …$$
10. 2, −5, 10, −17, 26, …
11. 3, 12, 48, 192, 768, …
12. $$\frac{5}{3}, \frac{5}{2}, 3, \frac{10}{3}, \frac{25}{7}, …$$
13. Write a recursive rule for 5, −15, 45, −135, 405, …
14. Simplify the factorial expression.
15. $$\frac{5!}{7!}$$
16. $$\frac{(n+1)!}{(n-1)!}$$
17. $$\frac{6!·n!}{3!·(n-1)!}$$
18. Problem Solving
19. If you put 1 grain of rice on the first square on a chess board, 2 grains on the second square, 4 grains on the third square, and continue to double. How many grains will be on the 64th square? Write a rule for the nth term of the sequence and then find the 64th term.
20. Mixed Review
21. (9-06) Use a matrix to find the equation of the line through (−3, 5) and (4, −2).
22. (9-05) Find cofactor C32: $$\left[\begin{matrix} 3 & -1 & 2 \\ 0 & 2 & 1 \\ 5 & -4 & -3 \end{matrix}\right]$$
23. (9-04) Find the inverse matrix: $$\left[\begin{matrix} 3 & 2 \\ 0 & 1 \end{matrix}\right]$$
24. (8-01) Solve by substitution: \left\{\begin{align} y &= 4x^2 \\ y &= 3x \end{align}\right.
25. (7-06) Graph the parametric equation: \left\{\begin{align} x &= 3t^2 \\ y &= \sqrt{t} \end{align}\right.

1. −1, −3, −5, −7, −9
2. 2, 6, 12, 20, 30
3. $$-\frac{1}{3}, \frac{1}{2}, -\frac{3}{5}, \frac{2}{3}, -\frac{5}{7}$$
4. 2, 7, 17, 37, 77
5. −3, 9, 81, 6561, 43046721
6. an = 4n − 3
7. $$a_n = \frac{2n-1}{2^n}$$
8. an = (−1)n−1·(n2 + 1)
9. an = 3·4n−1
10. $$a_n = \frac{5n}{n+2}$$
11. a1 = 5, an = −3an−1
12. $$\frac{1}{42}$$
13. (n + 1)n
14. 120n
15. an = 2n−1; 9.22×1018 grains
16. x + y = 2
17. −3
18. $$\left[\begin{matrix} \frac{1}{3} & -\frac{2}{3} \\ 0 & 1 \end{matrix}\right]$$
19. (0, 0), $$\left(\frac{3}{4}, \frac{9}{4}\right)$$