Precalculus by Richard Wright

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

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 \(a_n = 2^{n-1}\), where *n* is the domain (like *x*) and *a _{n}* is the range (like

- Plug in numbers for
*n*, usually 1, 2, 3, 4, … - This gives
*a*_{1},*a*_{2},*a*_{3},*a*_{4}, …

Write the first five terms of the sequence \(a_n = 3(-1)^{n} - n\).

Let *n* = 1.

$$ a_\color{blue}{1} = 3(-1)^{\color{blue}{1}} - \color{blue}{1} $$

$$ a_1 = -4 $$

Continue plugging in numbers for *n*.

$$ a_\color{blue}{2} = 3(-1)^{\color{blue}{2}} - \color{blue}{2} = 1 $$

$$ a_\color{red}{3} = 3(-1)^{\color{red}{3}} - \color{red}{3} = -6 $$

$$ a_\color{purple}{4} = 3(-1)^{\color{purple}{4}} - \color{purple}{4} = -1 $$

$$ a_\color{green}{5} = 3(-1)^{\color{green}{5}} - \color{green}{5} = -8 $$

The sequence is −4, 1, −6, −1, −8, …

Find the first five terms of \(a_n = 3n + 1\).

4, 7, 10, 13, 16

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
*a*_{1}is negative or \((-1)^{n-1}\) is*a*_{1}is positive. - Write a formula for
*a*in terms of_{n}*n*. Test your formula for*n*= 1,*n*= 2, and*n*= 3.

Find the rule for the *n*^{th} term of 3, 5, 7, 9, 11, …

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 *a _{n}*. Compare them to the

n |
1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

a_{n} |
\(3 + 2·0\), | \(3 + 2·1\), | \(3 + 2·2\), | \(3 + 2·3\), | \(3 + 2·4\), … |

Notice the multiples of 2 are 1 less than *n*. So, the rule for the *n*^{th} term is

$$ a_n = 3 + 2·(n-1) $$

Distribute and simplify to get

$$ a_n = 2n + 1 $$

Find the rule for the *n*^{th} term of 3, −6, 12, −24, 48, …

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, … |

\(3·2^0\), | \(-3·2^1\), | \(3·2^2\), | \(-3·2^3\), | \(3·2^4\), … |

These expressions are the *a _{n}*. Compare them to the

n |
1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

a_{n} |
\(3·2^0\) | \(-3·2^1\) | \(3·2^2\) | \(-3·2^3\) | \(3·2^4\) |

Notice the exponents are 1 less than *n*. There is also a pattern of alternating signs with *a*_{1} being positive so use \((-1)^{n-1}\). So, the rule for the *n*^{th} term is

$$ a_n = 3·2^{n-1}·(-1)^{n-1} $$

Find the rule for the *n*^{th} term of \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, …\)

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 *a _{n}* to the

n |
1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

a_{n} |
\(\frac{1}{2}\) | \(\frac{2}{3}\) | \(\frac{3}{4}\) | \(\frac{4}{5}\) | \(\frac{5}{6}\) |

Notice the numerators are equal to *n*. The denominators are 1 more than *n*. So, the rule for the *n*^{th} term is

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

Find the rule for the *n*^{th} term of 0, 3, 8, 15, 24, …

There is no constant pattern or adding or multiplying to get the next number. Start by comparing the the *a _{n}* to the

n |
1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

a_{n} |
0 | 3 | 8 | 15 | 24 |

Notice that the terms are close to perfect squares.

n |
1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

a_{n} |
0 | 3 | 8 | 15 | 24 |

a + 1_{n} |
0 + 1 = 1 | 3 + 1 = 4 | 8 + 1 = 9 | 15 + 1 = 16 | 24 + 1 = 25 |

Notice the \(a_n + 1 = n^2\). So, the rule for the *n*^{th} term is

$$ a_n = n^2 - 1 $$

Find the rule for the *n*^{th} term.

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

a. \(4n - 1\), b. \(\frac{2n}{2^{n-1}}\), c. \(n^3\)

Explicit rules give the term, *a _{n}*, directly by simply plugging in the

For a recursive rule, *a _{n}* refers to the current term.

Write the first five terms of \(a_1 = 4, a_{n} = 2a_{n-1} - 1\).

*a*_{1} is already given, so start by plugging in *n* = 2.

$$ a_1 = \color{red}{4} $$

$$ a_{n} = 2a_{n-1} - 1 $$

$$ a_{\color{blue}{2}} = 2a_\color{blue}{2-1} - 1 $$

$$ a_{\color{blue}{2}} = 2a_\color{blue}{1} - 1 $$

*a*_{1} is 4.

$$ a_\color{blue}{2} = 2(\color{blue}{4}) - 1 = \color{red}{7} $$

Plug in *n* = 3.

$$ a_{\color{blue}{3}} = 2a_\color{blue}{3-1} - 1 $$

$$ a_{\color{blue}{3}} = 2a_\color{blue}{2} - 1 $$

$$ a_\color{blue}{3} = 2(\color{blue}{7}) - 1 = \color{red}{13} $$

Plug in *n* = 4.

$$ a_{\color{blue}{4}} = 2a_\color{blue}{4-1} - 1 $$

$$ a_{\color{blue}{4}} = 2a_\color{blue}{3} - 1 $$

$$ a_\color{blue}{4} = 2(\color{blue}{13}) - 1 = \color{red}{25} $$

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

$$ a_\color{blue}{5} = 2(\color{blue}{25}) - 1 = \color{red}{49} $$

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

*a _{n}* refers to the current term.

- Write an expression for the term based on the previous term such as

$$ a_{n} = 2a_{n-1} - 1 $$ - Write the first term so we know where to start such as

$$ a_1 = 4 $$ - Put those together

$$ a_1 = 4, a_{n} = 2a_{n-1} - 1 $$

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

The pattern is adding 2.

$$ a_{n} = a_{n-1} + 2 $$

Don't forget the first term.

$$ a_n = 3, a_{n} = a_{n-1} + 2 $$

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

\(a_1 = 2, a_{n} = 3a_{n-1}\)

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.

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

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 \(\frac{2!3!}{4!}\)

\(\frac{1}{2}\)

- Plug in numbers for
*n*, usually 1, 2, 3, 4, … - This gives
*a*_{1},*a*_{2},*a*_{3},*a*_{4}, …

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
*a*_{1}is negative or \((-1)^{n-1}\) is*a*_{1}is positive. - Write a formula for
*a*in terms of_{n}*n*. Test your formula for*n*= 1,*n*= 2, and*n*= 3.

*a _{n}* refers to the current term.

- Write an expression for the term based on the previous term such as

$$ a_{n} = 2a_{n-1} - 1 $$ - Write the first term so we know where to start such as

$$ a_1 = 4 $$ - Put those together

$$ a_1 = 4, a_{n} = 2a_{n-1} - 1 $$

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.

Helpful videos about this lesson.

- \(a_n = -2n + 1\)
- \(a_n = n^2 + n\)
- \(a_n = (-1)^n \left(\frac{n}{n+2}\right)\)
- \(a_1 = 2, a_{n} = 2a_{n-1} + 3\)
- \(a_1 = -3, a_{n} = \left(a_{n-1}\right)^2\)
- 1, 5, 9, 13, 17, …
- \(\frac{1}{2}, \frac{3}{4}, \frac{5}{8}, \frac{7}{16}, \frac{9}{32}, …\)
- 2, −5, 10, −17, 26, …
- 3, 12, 48, 192, 768, …
- \(\frac{5}{3}, \frac{5}{2}, 3, \frac{10}{3}, \frac{25}{7}, …\)
- Write a recursive rule for 5, −15, 45, −135, 405, …
- \(\frac{5!}{7!}\)
- \(\frac{(n+1)!}{(n-1)!}\)
- \(\frac{6!·n!}{3!·(n-1)!}\)
- 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 64
^{th}square? Write a rule for the*n*^{th}term of the sequence and then find the 64^{th}term. - (9-06) Use a matrix to find the equation of the line through (−3, 5) and (4, −2).
- (9-05) Find cofactor
*C*_{32}: \(\left[\begin{matrix} 3 & -1 & 2 \\ 0 & 2 & 1 \\ 5 & -4 & -3 \end{matrix}\right]\) - (9-04) Find the inverse matrix: \(\left[\begin{matrix} 3 & 2 \\ 0 & 1 \end{matrix}\right]\)
- (8-01) Solve by substitution: \(\left\{\begin{align} y &= 4x^2 \\ y &= 3x \end{align}\right.\)
- (7-06) Graph the parametric equation: \(\left\{\begin{align} x &= 3t^2 \\ y &= \sqrt{t} \end{align}\right.\)

Write the first 5 term of the sequence.

Write the rule for the *n*^{th} term.

Simplify the factorial expression.

Problem Solving

Mixed Review

- −1, −3, −5, −7, −9
- 2, 6, 12, 20, 30
- \(-\frac{1}{3}, \frac{1}{2}, -\frac{3}{5}, \frac{2}{3}, -\frac{5}{7}\)
- 2, 7, 17, 37, 77
- −3, 9, 81, 6561, 43046721
- \(a_{n} = 4n - 3\)
- \(a_n = \frac{2n-1}{2^n}\)
- \(a_n = (-1)^{n-1}·\left(n^2 + 1\right)\)
- \(a_n = 3·4^{n-1}\)
- \(a_n = \frac{5n}{n+2}\)
- \(a_1 = 5, a_{n+1} = -3a_n\)
- \(\frac{1}{42}\)
- \((n + 1)n\)
- \(120n\)
- \(a_n = 2^{n-1}; 9.22×10^{18}\) grains
- \(x + y = 2\)
- −3
- \(\left[\begin{matrix} \frac{1}{3} & -\frac{2}{3} \\ 0 & 1 \end{matrix}\right]\)
- (0, 0), \(\left(\frac{3}{4}, \frac{9}{4}\right)\)