Chapter 13

A sequence is an ordered list of numbers that can be either finite or infinite in number. When a finite sequence is defined by a formula, its domain is a subset of the non-negative integers. When an infinite sequence is defined by a formula, its domain is all positive or all non-negative integers.

Yes, both sets go on indefinitely, so they are both infinite sequences.

A factorial is the product of a positive integer and all the positive integers below it. An exclamation point is used to indicate the operation. Answers may vary. An example of the benefit of using factorial notation is when indicating the product It is much easier to write than it is to write out $\text{13}\cdot \text{12}\cdot \text{11}\cdot \text{10}\cdot \text{9}\cdot \text{8}\cdot \text{7}\cdot \text{6}\cdot \text{5}\cdot \text{4}\cdot \text{3}\cdot \text{2}\cdot \text{1}\text{.}$

First four terms: $-8,-\frac{16}{3},-4,-\frac{16}{5}$

First four terms: $2,\frac{1}{2},\frac{8}{27},\frac{1}{4}$ .

First four terms: $1.25,-5,20,-80$ .

First four terms: $\frac{1}{3},\frac{4}{5},\frac{9}{7},\frac{16}{9}$ .

First four terms: $-\frac{4}{5},4,-20,100$

$\frac{1}{3},\frac{4}{5},\frac{9}{7},\frac{16}{9},\frac{25}{11},31,44,59$

$-0.6,-3,-15,-20,-375,-80,-9375,-320$

$a_n=n^2+3$

$a_n=\frac{2^n}{2n}\text{or }\frac{2^{n-1}}{n}$

$a_n={\left(-\frac{1}{2}\right)}^{n-1}$

First five terms: $3,-9,27,-81,243$

First five terms: $-1,1,-9,\frac{27}{11},\frac{891}{5}$

$\frac{1}{24},\text{1, }\frac{1}{4},\frac{3}{2},\frac{9}{4},\frac{81}{4},\frac{2187}{8},\frac{531,441}{16}$

$2,10,12,\frac{14}{5},\frac{4}{5},2,10,12$

$a_1=-8,a_n=a_{n-1}+n$

$a_1=35,a_n=a_{n-1}+3$

$720$

$665,280$

First four terms: $1,\frac{1}{2},\frac{2}{3},\frac{3}{2}$

First four terms: $-1,2,\frac{6}{5},\frac{24}{11}$

$a_n=2^{n-2}$

$a_1=6,a_n=2a_{n-1}-5$

First five terms: $\frac{29}{37}$, $\frac{152}{111}$, $\frac{716}{333}$, $\frac{3188}{999}$, $\frac{13724}{2997}$

First five terms: 2, 3, 5, 17, 65537

$a_{10}=7,257,600$

First six terms: 0.042, 0.146, 0.875, 2.385, 4.708

First four terms: 5.975, 2.765, 185.743, 1057.25, 6023.521

If $a_n=-421$ is a term in the sequence, then solving the equation $-421=-6-8n$ for $n$ will yield a non-negative integer. However, if $-421=-6-8n,$ then $n=51.875$ so $a_n=-421$ is not a term in the sequence.

$a_1=1,a_2=0,a_n=a_{n-1}-a_{n-2}$

$\frac{(n+2)!}{(n-1)!}=\frac{(n+2)·(n+1)·(n)·(n-1)·\mathrm{...}·3·2·1}{(n-1)·\mathrm{...}·3·2·1}=n(n+1)(n+2)=n^3+3n^2+2n$

A sequence where each successive term of the sequence increases (or decreases) by a constant value.

We find whether the difference between all consecutive terms is the same. This is the same as saying that the sequence has a common difference.

Both arithmetic sequences and linear functions have a constant rate of change. They are different because their domains are not the same; linear functions are defined for all real numbers, and arithmetic sequences are defined for natural numbers or a subset of the natural numbers.

The common difference is $\frac{1}{2}$

The sequence is not arithmetic because $16-4\ne 64-16.$

$0,\frac{2}{3},\frac{4}{3},2,\frac{8}{3}$

$0,-5,-10,-15,-20$

$a_4=19$

$a_6=41$

$a_1=2$

$a_1=5$

$a_1=6$

$a_{21}=-13.5$

$-19,-20.4,-21.8,-23.2,-24.6$

$\begin{array}{ll}{a}_1=17; a_n=a_{n-1}+9\hfill & n\ge 2\hfill \end{array}$

$\begin{array}{ll}{a}_1=12; a_n=a_{n-1}+5\hfill & n\ge 2\hfill \end{array}$

$\begin{array}{ll}{a}_1=8.9; a_n=a_{n-1}+1.4\hfill & n\ge 2\hfill \end{array}$

$\begin{array}{ll}{a}_1=\frac{1}{5}; a_n=a_{n-1}+\frac{1}{4}\hfill & n\ge 2\hfill \end{array}$

$\begin{array}{ll}{}_1=\frac{1}{6}; a_n=a_{n-1}-\frac{13}{12}\hfill & n\ge 2\hfill \end{array}$

$a_1=4;a_n=a_{n-1}+7;a_{14}=95$

First five terms: $20,16,12,8,4.$

$a_n=1+2n$

$a_n=-105+100n$

$a_n=1.8n$

$a_n=13.1+2.7n$

$a_n=\frac{1}{3}n-\frac{1}{3}$

There are 10 terms in the sequence.

There are 6 terms in the sequence.

The graph does not represent an arithmetic sequence.

$1,4,7,10,13,16,19$

Answers will vary. Examples: $a_n=20.6n$ and $a_n=2+20.4\mathrm{n.}$

$a_{11}=-17a+38b$

The sequence begins to have negative values at the 13^th term, $a_{13}=-\frac{1}{3}$

Answers will vary. Check to see that the sequence is arithmetic. Example: Recursive formula: $a_1=3,a_n=a_{n-1}-3.$ First 4 terms: $\begin{array}{ll}3,0,-3,-6\hfill & a_{31}=-87\hfill \end{array}$

A sequence in which the ratio between any two consecutive terms is constant.

Divide each term in a sequence by the preceding term. If the resulting quotients are equal, then the sequence is geometric.

Both geometric sequences and exponential functions have a constant ratio. However, their domains are not the same. Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.

The common ratio is $-2$

The sequence is geometric. The common ratio is 2.

The sequence is geometric. The common ratio is $-\frac{1}{2}.$

The sequence is geometric. The common ratio is $5.$

$5,1,\frac{1}{5},\frac{1}{25},\frac{1}{125}$

$800,400,200,100,50$

$a_4=-\frac{16}{27}$

$a_7=-\frac{2}{729}$

$7,1.4,0.28,0.056,0.0112$

$\begin{array}{cc}a{}_1=-32,& a_n=\frac{1}{2}{a}_{n-1}\end{array}$

$\begin{array}{cc}{a}_1=10,& a_n=-0.3a_{n-1}\end{array}$

$\begin{array}{cc}{a}_1=\frac{3}{5},& a_n=\frac{1}{6}{a}_{n-1}\end{array}$

$a_1=\frac{1}{512},a_n=-4a_{n-1}$

$12,-6,3,-\frac{3}{2},\frac{3}{4}$

$a_n=3^{n-1}$

$a_n=0.8\cdot {(-5)}^{n-1}$

$a_n=-{\left(\frac{4}{5}\right)}^{n-1}$

$a_n=3\cdot {\left(-\frac{1}{3}\right)}^{n-1}$

$a_{12}=\frac{1}{177,147}$

There are $12$ terms in the sequence.

The graph does not represent a geometric sequence.

Answers will vary. Examples: ${\begin{array}{cc}{a}_1=800,& a_n=0.5a\end{array}}_{n-1}$ and ${\begin{array}{cc}{a}_1=12.5,& a_n=4a\end{array}}_{n-1}$

$a_5=256b$

The sequence exceeds $100$ at the 14^th term, $a_{14}\approx 107.$

$a_4=-\frac{32}{3}$ is the first non-integer value

Answers will vary. Example: Explicit formula with a decimal common ratio: $a_n=400\cdot {0.5}^{n-1};$ First 4 terms: $\begin{array}{cc}400,200,100,50;& a_8=3.125\end{array}$

An $n\text{th}$ partial sum is the sum of the first $n$ terms of a sequence.

A geometric series is the sum of the terms in a geometric sequence.

An annuity is a series of regular equal payments that earn a constant compounded interest.

${\displaystyle \sum _{n=0}^{4}5n}$

${\displaystyle \sum _{k=1}^{5}4}$

${\displaystyle \sum _{k=1}^{20}8k+2}$

$S_5=\frac{5\left(\frac{3}{2}+\frac{7}{2}\right)}{2}$

$S_{13}=\frac{13\left(3.2+5.6\right)}{2}$

${\displaystyle \sum _{k=1}^{7}8\cdot {0.5}^{k-1}}$

$S_5=\frac{9\left(1-{\left(\frac{1}{3}\right)}^5\right)}{1-\frac{1}{3}}=\frac{121}{9}\approx 13.44$

$S_{11}=\frac{64\left(1-{0.2}^{11}\right)}{1-0.2}=\frac{781,249,984}{9,765,625}\approx 80$

The series is defined. $S=\frac{2}{1-0.8}$

The series is defined. $S=\frac{-1}{1-\left(-\frac{1}{2}\right)}$

Sample answer: The graph of $S_n$ seems to be approaching 1. This makes sense because ${\displaystyle \sum _{k=1}^{\infty }{\left(\frac{1}{2}\right)}^k}$ is a defined infinite geometric series with $S=\frac{\frac{1}{2}}{1–\left(\frac{1}{2}\right)}=1.$

49

254

$S_7=\frac{147}{2}$

$S_{11}=\frac{55}{2}$

$S_7=5208.4$

$S_{10}=-\frac{1023}{256}$

$S=-\frac{4}{3}$

$S=9.2$

$3,705.42

$695,823.97

$a_k=30-k$

9 terms

$r=\frac{4}{5}$

$400 per month

420 feet

12 feet

There are $m+n$ ways for either event $A$ or event $B$ to occur.

The addition principle is applied when determining the total possible of outcomes of either event occurring. The multiplication principle is applied when determining the total possible outcomes of both events occurring. The word “or” usually implies an addition problem. The word “and” usually implies a multiplication problem.

A combination; $C(n,r)=\frac{n!}{(n-r)!r!}$

$4+2=6$

$5+4+7=16$

$2\times 6=12$

${10}^3=1000$

$P(5,2)=20$

$P(3,3)=6$

$P(11,5)=55,440$

$C(12,4)=495$

$C(7,6)=7$

$2^{10}=1024$

$2^{12}=4096$

$2^9=512$

$\frac{8!}{3!}=6720$

$\frac{12!}{3!2!3!4!}$

9

Yes, for the trivial cases $r=0$ and $r=1.$ If $r=0,$ then $C(n,r)=P(n,r)=1\text{.\hspace{0.17em}}$ If $r=1,$ then $r=1,$ $C(n,r)=P(n,r)=n.$

$\frac{6!}{2!}\times 4!=8640$

$6-3+8-3=8$

$4\times 2\times 5=40$

$4\times 12\times 3=144$

$P(15,9)=1,816,214,400$

$C(10,3)\times C(6,5)\times C(5,2)=7,200$

$2^{11}=2048$

$\frac{20!}{6!6!8!}=116,396,280$

A binomial coefficient is an alternative way of denoting the combination $C(n,r).$ It is defined as $\left(\begin{array}{c}n\\ r\end{array}\right)=C(n,r)=\frac{n!}{r!(n-r)!}.$

The Binomial Theorem is defined as ${(x+y)}^n={\displaystyle \sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^{n-k}}{y}^k$ and can be used to expand any binomial.

15

35

10

12,376

$64a^3-48a^{2}b+12a{b}^2-b^3$

$27a^3+54a^{2}b+36a{b}^2+8b^3$

$1024x^5+2560x^{4}y+2560x^3{y}^2+1280x^2{y}^3+320x{y}^4+32y^5$

$1024x^5-3840x^{4}y+5760x^3{y}^2-4320x^2{y}^3+1620x{y}^4-243y^5$

$\frac{1}{x^4}+\frac{8}{x^{3}y}+\frac{24}{x^2{y}^2}+\frac{32}{x{y}^3}+\frac{16}{y^4}$