第13章数列、確率、数え上げSequences, Probability, and Counting Theory

Answer Key解答

Solution 1解答 1

The first five terms are ${1, 6, 11, 16, 21} .$

初めの五項は ${1, 6, 11, 16, 21}$ である。

Solution 2解答 2

The first five terms are ${- 2, 2, - \frac{3}{2}, 1, - \frac{5}{8}} .$

初めの五項は ${- 2, 2, - \frac{3}{2}, 1, - \frac{5}{8}}$ である。

Solution 3解答 3

The first six terms are ${2, 5, 54, 10, 250, 15} .$

初めの六項は ${2, 5, 54, 10, 250, 15}$ である。

Solution 4解答 4

$a_{n} = {(- 1)}^{n + 1} 9^{n}$

Solution 5解答 5

$a_{n} = - \frac{3^{n}}{4 n}$

Solution 6解答 6

$a_{n} = e^{n - 3}$

Solution 7解答 7

${2, 5, 11, 23, 47}$

Solution 8解答 8

${0, 1, 1, 1, 2, 3, \frac{5}{2}, \frac{17}{6}} .$

Solution 9解答 9

The first five terms are ${1, \frac{3}{2}, 4, 15, 72} .$

初めの五項は ${1, \frac{3}{2}, 4, 15, 72}$ である。

Solution 1解答 1

The sequence is arithmetic. The common difference is $- 2.$

この数列は等差数列である。公差は $- 2.$ である。

Solution 2解答 2

The sequence is not arithmetic because $3 - 1 \neq 6 - 3.$

$3 - 1 \neq 6 - 3.$ なので、この数列は等差数列でない。

Solution 3解答 3

${1, 6, 11, 16, 21}$

Solution 4解答 4

$a_{2} = 2$

Solution 5解答 5

$\begin{array}{l} a_{1} = 25 \\ a_{n} = a_{n - 1} + 12, for n \geq 2 \end{array}$

式中の英語：for：ただし

Solution 6解答 6

$a_{n} = 53 - 3 n$

Solution 7解答 7

There are 11 terms in the sequence.

この数列には11項ある。

Solution 8解答 8

The formula is $T_{n} = 10 + 4 n,$ and it will take her 42 minutes.

公式は $T_{n} = 10 + 4 n$ であり、彼女は42分かかる。

Solution 1解答 1

The sequence is not geometric because $\frac{10}{5} \neq \frac{15}{10}$ .

$\frac{10}{5} \neq \frac{15}{10}$ なので、この数列は等比数列でない。

Solution 2解答 2

The sequence is geometric. The common ratio is $\frac{1}{5}$ .

この数列は等比数列である。公比は $\frac{1}{5}$ である。

Solution 3解答 3

${18, 6, 2, \frac{2}{3}, \frac{2}{9}}$

Solution 4解答 4

$\begin{array}{l} a_{1} = 2 \\ a_{n} = \frac{2}{3} a_{n - 1} for n \geq 2 \end{array}$

式中の英語：for：ただし

Solution 5解答 5

$a_{6} = 16, 384$

Solution 6解答 6

$a_{n} = - {(- 3)}^{n - 1}$

Solution 7解答 7

ⓐ $P_{n} = 293 \cdot 1.026 a^{n}$ ⓑThe number of hits will be about 333.

ⓑ閲覧数はおよそ333になる。

Solution 1解答 1

Solution 2解答 2

$26 .4$

Solution 3解答 3

$328$

Solution 4解答 4

$−280$

Solution 5解答 5

$2,025

Solution 6解答 6

$\approx 2, 000.00$

Solution 7解答 7

9,840

Solution 8解答 8

$275,513.31

Solution 9解答 9

The sum is not defined.

和は定められない。

Solution 10解答 10

The sum of the infinite series is defined.

この無限級数の和は定められる。

Solution 11解答 11

The sum of the infinite series is defined.

この無限級数の和は定められる。

Solution 12解答 12

Solution 13解答 13

The series is not geometric.

この級数は等比級数でない。

Solution 14解答 14

$- \frac{3}{11}$

Solution 15解答 15

$32,775.87

Solution 1解答 1

Solution 2解答 2

There are 60 possible breakfast specials.

朝食の特別献立は60通りありうる。

Solution 3解答 3

120

Solution 4解答 4

Solution 5解答 5

Solution 6解答 6

$P (7, 7) = 5, 040$

Solution 7解答 7

$P (7, 5) = 2, 520$

Solution 8解答 8

$C (10, 3) = 120$

Solution 9解答 9

64 sundaes

サンデー64通り

Solution 10解答 10

840

Solution 1解答 1

ⓐ35ⓑ330

Solution 2解答 2

ⓐ $x^{5} - 5 x^{4} y + 10 x^{3} y^{2} - 10 x^{2} y^{3} + 5 x y^{4} - y^{5}$ ⓑ $8 x^{3} + 60 x^{2} y + 150 x y^{2} + 125 y^{3}$

Solution 3解答 3

$- 10, 206 x^{4} y^{5}$

Solution 1解答 1

Outcome結果	Probability確率
Heads表	$\frac{1}{2}$
Tails裏	$\frac{1}{2}$

Solution 2解答 2

$\frac{2}{3}$

Solution 3解答 3

$\frac{7}{13}$

Solution 4解答 4

$\frac{2}{13}$

Solution 5解答 5

$\frac{5}{6}$

Solution 6解答 6

$\begin{array}{l} a . \frac{1}{91}; & b . \frac{5}{91}; & c . \frac{86}{91} \end{array}$

13.1 Section Exercises13.1 節末問題

Solution 1解答 1

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.

数列とは、有限にも無限にもなりうる、順序を持つ数の並びである。有限数列が公式で定められるとき、その定義域は非負の整数の部分集合である。無限数列が公式で定められるとき、その定義域は正の整数すべて、または非負の整数すべてである。

Solution 3解答 3

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

そうである。どちらの集合も限りなく続くので、どちらも無限数列である。

Solution 5解答 5

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 $13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 .$

階乗とは、ある正の整数と、それより下のすべての正の整数との積である。この演算を示すのに感嘆符を使う。答えはいろいろありうる。階乗の表記の利点の例は、積を示すときである。 $13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ と書き出すよりずっと楽に書ける。

Solution 7解答 7

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

初めの四項: $- 8, - \frac{16}{3}, - 4, - \frac{16}{5}$

Solution 9解答 9

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

初めの四項: $2, \frac{1}{2}, \frac{8}{27}, \frac{1}{4}$ 。

Solution 11解答 11

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

初めの四項: $1.25, - 5, 20, - 80$ 。

Solution 13解答 13

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

初めの四項: $\frac{1}{3}, \frac{4}{5}, \frac{9}{7}, \frac{16}{9}$ 。

Solution 15解答 15

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

初めの四項: $- \frac{4}{5}, 4, - 20, 100$

Solution 17解答 17

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

Solution 19解答 19

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

Solution 21解答 21

$a_{n} = n^{2} + 3$

Solution 23解答 23

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

式中の英語：or：または

Solution 25解答 25

$a_{n} = {(- \frac{1}{2})}^{n - 1}$

Solution 27解答 27

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

初めの五項: $3, - 9, 27, - 81, 243$

Solution 29解答 29

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

初めの五項: $- 1, 1, - 9, \frac{27}{11}, \frac{891}{5}$

Solution 31解答 31

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

Solution 33解答 33

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

Solution 35解答 35

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

Solution 37解答 37

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

Solution 39解答 39

$720$

Solution 41解答 41

$665, 280$

Solution 43解答 43

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

初めの四項: $1, \frac{1}{2}, \frac{2}{3}, \frac{3}{2}$

Solution 45解答 45

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

初めの四項: $- 1, 2, \frac{6}{5}, \frac{24}{11}$

Solution 47解答 47

Graph of a scattered plot with points at (1, 0), (2, 5/2), (3, 8/3), (4, 17/4), and (5, 24/5). The x-axis is labeled n and the y-axis is labeled a_n.

Solution 49解答 49

Graph of a scattered plot with points at (1, 2), (2, 1), (3, 0), (4, 1), and (5, 0). The x-axis is labeled n and the y-axis is labeled a_n.

Solution 51解答 51

Graph of a scattered plot with labeled points: (1, 2), (2, 6), (3, 12), (4, 20), and (5, 30). The x-axis is labeled n and the y-axis is labeled a_n.

Solution 53解答 53

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

Solution 55解答 55

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

Solution 57解答 57

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

初めの五項: $\frac{29}{37}$ 、 $\frac{152}{111}$ 、 $\frac{716}{333}$ 、 $\frac{3188}{999}$ 、 $\frac{13724}{2997}$

Solution 59解答 59

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

初めの五項: 2, 3, 5, 17, 65537

Solution 61解答 61

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

Solution 63解答 63

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

初めの六項: 0.042, 0.146, 0.875, 2.385, 4.708

Solution 65解答 65

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

初めの四項: 5.975, 2.765, 185.743, 1057.25, 6023.521

Solution 67解答 67

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

$a_{n} = - 421$ がこの数列の項なら、式 $- 421 = - 6 - 8 n$ を $n$ について解くと非負の整数になる。だが $- 421 = - 6 - 8 n$ なら $n = 51.875$ なので、 $a_{n} = - 421$ はこの数列の項でない。

Solution 69解答 69

$a_{1} = 1, a_{2} = 0, a_{n} = a_{n - 1} - a_{n - 2}$

Solution 71解答 71

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

13.2 Section Exercises13.2 節末問題

Solution 1解答 1

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

後に続く各項が一定の値だけ増える(または減る)数列。

Solution 3解答 3

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.

連続するすべての項の差が同じかどうかを調べる。これは、その数列が公差を持つと言うのと同じである。

Solution 5解答 5

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.

等差数列も一次関数も変化の割合が一定である。違うのは定義域が同じでない点である。一次関数はすべての実数について定められ、等差数列は自然数またはその部分集合について定められる。

Solution 7解答 7

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

公差は $\frac{1}{2}$ である。

Solution 9解答 9

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

$16 - 4 \neq 64 - 16.$ なので、この数列は等差数列でない。

Solution 11解答 11

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

Solution 13解答 13

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

Solution 15解答 15

$a_{4} = 19$

Solution 17解答 17

$a_{6} = 41$

Solution 19解答 19

$a_{1} = 2$

Solution 21解答 21

$a_{1} = 5$

Solution 23解答 23

$a_{1} = 6$

Solution 25解答 25

$a_{21} = - 13.5$

Solution 27解答 27

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

Solution 29解答 29

$\begin{array}{l} a_{1} = 17; a_{n} = a_{n - 1} + 9 & n \geq 2 \end{array}$

Solution 31解答 31

$\begin{array}{l} a_{1} = 12; a_{n} = a_{n - 1} + 5 & n \geq 2 \end{array}$

Solution 33解答 33

$\begin{array}{l} a_{1} = 8.9; a_{n} = a_{n - 1} + 1.4 & n \geq 2 \end{array}$

Solution 35解答 35

$\begin{array}{l} a_{1} = \frac{1}{5}; a_{n} = a_{n - 1} + \frac{1}{4} & n \geq 2 \end{array}$

Solution 37解答 37

$\begin{array}{l} _{1} = \frac{1}{6}; a_{n} = a_{n - 1} - \frac{13}{12} & n \geq 2 \end{array}$

Solution 39解答 39

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

Solution 41解答 41

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

初めの五項: $20, 16, 12, 8, 4.$

Solution 43解答 43

$a_{n} = 1 + 2 n$

Solution 45解答 45

$a_{n} = - 105 + 100 n$

Solution 47解答 47

$a_{n} = 1.8 n$

Solution 49解答 49

$a_{n} = 13.1 + 2.7 n$

Solution 51解答 51

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

Solution 53解答 53

There are 10 terms in the sequence.

この数列には10項ある。

Solution 55解答 55

There are 6 terms in the sequence.

この数列には6項ある。

Solution 57解答 57

The graph does not represent an arithmetic sequence.

このグラフは等差数列を表さない。

Solution 59解答 59

Graph of a scattered plot with labeled points: (1, 9), (2, -1), (3, -11), (4, -21), and (5, -31). The x-axis is labeled n and the y-axis is labeled a_n.

Solution 61解答 61

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

Solution 63解答 63

Graph of a scattered plot with labeled points: (1, 1), (2, 4), (3, 7), (4, 10), and (5, 13). The x-axis is labeled n and the y-axis is labeled a_n.

Solution 65解答 65

Graph of a scattered plot with labeled points: (1, 5.5), (2, 6), (3, 6.5), (4, 7), and (5, 7.5). The x-axis is labeled n and the y-axis is labeled a_n.

Solution 67解答 67

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

答えはいろいろありうる。例: $a_{n} = 20.6 n$ と $a_{n} = 2 + 20.4 n.$

Solution 69解答 69

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

Solution 71解答 71

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

この数列は第13項 $a_{13} = - \frac{1}{3}$ から負の値を持ち始める。

Solution 73解答 73

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: $\begin{array}{l} 3, 0, - 3, - 6 & a_{31} = - 87 \end{array}$

答えはいろいろありうる。等差数列になっているか確かめよ。例: 漸化式 $a_{1} = 3, a_{n} = a_{n - 1} - 3.$ 初めの4項: $\begin{array}{l} 3, 0, - 3, - 6 & a_{31} = - 87 \end{array}$

13.3 Section Exercises13.3 節末問題

Solution 1解答 1

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

連続するどの二項の比も一定である数列。

Solution 3解答 3

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

数列の各項を直前の項で割る。できた商が等しいなら、その数列は等比数列である。

Solution 5解答 5

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.

等比数列も指数関数も一定の比を持つ。だが定義域は同じでない。指数関数はすべての実数について定められ、等比数列は正の整数についてだけ定められる。もう一つの違いは、等比数列の底(公比)は負でありうるが、指数関数の底は正でなければならない点である。

Solution 7解答 7

The common ratio is $- 2$

公比は $- 2$ である。

Solution 9解答 9

The sequence is geometric. The common ratio is 2.

この数列は等比数列である。公比は2である。

Solution 11解答 11

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

この数列は等比数列である。公比は $- \frac{1}{2}$ である。

Solution 13解答 13

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

この数列は等比数列である。公比は $5.$ である。

Solution 15解答 15

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

Solution 17解答 17

$800, 400, 200, 100, 50$

Solution 19解答 19

$a_{4} = - \frac{16}{27}$

Solution 21解答 21

$a_{7} = - \frac{2}{729}$

Solution 23解答 23

$7, 1.4, 0.28, 0.056, 0.0112$

Solution 25解答 25

$\begin{matrix} a {}_{1}= - 32, & a_{n} = \frac{1}{2} a_{n - 1} \end{matrix}$

Solution 27解答 27

$\begin{matrix} a_{1} = 10, & a_{n} = - 0.3 a_{n - 1} \end{matrix}$

Solution 29解答 29

$\begin{matrix} a_{1} = \frac{3}{5}, & a_{n} = \frac{1}{6} a_{n - 1} \end{matrix}$

Solution 31解答 31

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

Solution 33解答 33

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

Solution 35解答 35

$a_{n} = 3^{n - 1}$

Solution 37解答 37

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

Solution 39解答 39

$a_{n} = - {(\frac{4}{5})}^{n - 1}$

Solution 41解答 41

$a_{n} = 3 \cdot {(- \frac{1}{3})}^{n - 1}$

Solution 43解答 43

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

Solution 45解答 45

There are $12$ terms in the sequence.

この数列には $12$ 項ある。

Solution 47解答 47

The graph does not represent a geometric sequence.

このグラフは等比数列を表さない。

Solution 49解答 49

Graph of a scattered plot with labeled points: (1, 3), (2, 6), (3, 12), (4, 24), and (5, 48). The x-axis is labeled n and the y-axis is labeled a_n.

Solution 51解答 51

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

答えはいろいろありうる。例: ${\begin{matrix} a_{1} = 800, & a_{n} = 0.5 a \end{matrix}}_{n - 1}$ と ${\begin{matrix} a_{1} = 12.5, & a_{n} = 4 a \end{matrix}}_{n - 1}$

Solution 53解答 53

$a_{5} = 256 b$

Solution 55解答 55

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

この数列は第14項 $a_{14} \approx 107.$ で $100$ を超える。

Solution 57解答 57

$a_{4} = - \frac{32}{3}$ is the first non-integer value

$a_{4} = - \frac{32}{3}$ が初めて整数でない値になる

Solution 59解答 59

Answers will vary. Example: Explicit formula with a decimal common ratio: $a_{n} = 400 \cdot {0.5}^{n - 1};$ First 4 terms: $\begin{matrix} 400, 200, 100, 50; & a_{8} = 3.125 \end{matrix}$

答えはいろいろありうる。例: 公比が小数の一般項の公式: $a_{n} = 400 \cdot {0.5}^{n - 1};$ 初めの4項: $\begin{matrix} 400, 200, 100, 50; & a_{8} = 3.125 \end{matrix}$

13.4 Section Exercises13.4 節末問題

Solution 1解答 1

An $n th$ partial sum is the sum of the first $n$ terms of a sequence.

第 $n th$ 部分和とは、数列の初めの $n$ 項の和である。

Solution 3解答 3

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

等比級数とは、等比数列の項の和である。

Solution 5解答 5

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

年金とは、一定の複利がつく定期の等しい支払いの連なりである。

Solution 7解答 7

$\sum_{n = 0}^{4} 5 n$

Solution 9解答 9

$\sum_{k = 1}^{5} 4$

Solution 11解答 11

$\sum_{k = 1}^{20} 8 k + 2$

Solution 13解答 13

$S_{5} = \frac{5 (\frac{3}{2} + \frac{7}{2})}{2}$

Solution 15解答 15

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

Solution 17解答 17

$\sum_{k = 1}^{7} 8 \cdot {0.5}^{k - 1}$

Solution 19解答 19

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

Solution 21解答 21

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

Solution 23解答 23

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

この級数は定められる。 $S = \frac{2}{1 - 0.8}$

Solution 25解答 25

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

この級数は定められる。 $S = \frac{- 1}{1 - (- \frac{1}{2})}$

Solution 27解答 27

Graph of Javier's deposits where the x-axis is the months of the year and the y-axis is the sum of deposits.

Solution 29解答 29

Sample answer: The graph of $S_{n}$ seems to be approaching 1. This makes sense because $\sum_{k = 1}^{\infty} {(\frac{1}{2})}^{k}$ is a defined infinite geometric series with $S = \frac{\frac{1}{2}}{1 - (\frac{1}{2})} = 1.$

答えの例: $S_{n}$ のグラフは1に近づいているように見える。 $\sum_{k = 1}^{\infty} {(\frac{1}{2})}^{k}$ は $S = \frac{\frac{1}{2}}{1 - (\frac{1}{2})} = 1.$ である定められた無限等比級数なので、これは理にかなう。

Solution 31解答 31

Solution 33解答 33

254

Solution 35解答 35

$S_{7} = \frac{147}{2}$

Solution 37解答 37

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

Solution 39解答 39

$S_{7} = 5208.4$

Solution 41解答 41

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

Solution 43解答 43

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

Solution 45解答 45

$S = 9.2$

Solution 47解答 47

$3,705.42

Solution 49解答 49

$695,823.97

Solution 51解答 51

$a_{k} = 30 - k$

Solution 53解答 53

9 terms

9項

Solution 55解答 55

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

Solution 57解答 57

$400 per month

毎月400ドル

Solution 59解答 59

420 feet

420フィート

Solution 61解答 61

12 feet

12フィート

13.5 Section Exercises13.5 節末問題

Solution 1解答 1

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

出来事 $A$ か出来事 $B$ のどちらかが起こるしかたは $m + n$ 通りある。

Solution 3解答 3

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.

和の原理は、どちらかの出来事が起こるありうる結果の総数を決めるときに使う。積の原理は、両方の出来事が起こるありうる結果の総数を決めるときに使う。「または」という語はふつう足し算の問題を意味する。「かつ」という語はふつう掛け算の問題を意味する。

Solution 5解答 5

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

組合せ。 $C (n, r) = \frac{n!}{(n - r)! r!}$

Solution 7解答 7

$4 + 2 = 6$

Solution 9解答 9

$5 + 4 + 7 = 16$

Solution 11解答 11

$2 \times 6 = 12$

Solution 13解答 13

$10^{3} = 1000$

Solution 15解答 15

$P (5, 2) = 20$

Solution 17解答 17

$P (3, 3) = 6$

Solution 19解答 19

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

Solution 21解答 21

$C (12, 4) = 495$

Solution 23解答 23

$C (7, 6) = 7$

Solution 25解答 25

$2^{10} = 1024$

Solution 27解答 27

$2^{12} = 4096$

Solution 29解答 29

$2^{9} = 512$

Solution 31解答 31

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

Solution 33解答 33

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

Solution 35解答 35

Solution 37解答 37

Yes, for the trivial cases $r = 0$ and $r = 1.$ If $r = 0,$ then $C (n, r) = P (n, r) = 1 .$ If $r = 1,$ then $r = 1,$ $C (n, r) = P (n, r) = n .$

そうである。自明な場合 $r = 0$ と $r = 1.$ である。 $r = 0$ なら $C (n, r) = P (n, r) = 1$ である。 $r = 1$ なら $r = 1$ $C (n, r) = P (n, r) = n$ である。

Solution 39解答 39

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

Solution 41解答 41

$6 - 3 + 8 - 3 = 8$

Solution 43解答 43

$4 \times 2 \times 5 = 40$

Solution 45解答 45

$4 \times 12 \times 3 = 144$

Solution 47解答 47

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

Solution 49解答 49

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

Solution 51解答 51

$2^{11} = 2048$

Solution 53解答 53

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

13.6 Section Exercises13.6 節末問題

Solution 1解答 1

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

二項係数とは、組合せ $C (n, r).$ を表す別のしかたである。 $(\begin{matrix} n \\ r \end{matrix}) = C (n, r) = \frac{n!}{r! (n - r)!}$ と定められる。

Solution 3解答 3

The Binomial Theorem is defined as ${(x + y)}^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} y^{k}$ and can be used to expand any binomial.

二項定理は ${(x + y)}^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} y^{k}$ と定められ、どの二項式の展開にも使える。

Solution 5解答 5

Solution 7解答 7

Solution 9解答 9

Solution 11解答 11

12,376

Solution 13解答 13

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

Solution 15解答 15

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

Solution 17解答 17

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

Solution 19解答 19

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

Solution 21解答 21

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

Solution 23解答 23

$a^{17} + 17 a^{16} b + 136 a^{15} b^{2}$

Solution 25解答 25

$a^{15} - 30 a^{14} b + 420 a^{13} b^{2}$

Solution 27解答 27

$3, 486, 784, 401 a^{20} + 23, 245, 229, 340 a^{19} b + 73, 609, 892, 910 a^{18} b^{2}$

Solution 29解答 29

$x^{24} - 8 x^{21} \sqrt{y} + 28 x^{18} y$

Solution 31解答 31

$- 720 x^{2} y^{3}$

Solution 33解答 33

$220, 812, 466, 875, 000 y^{7}$

Solution 35解答 35

$35 x^{3} y^{4}$

Solution 37解答 37

$1, 082, 565 a^{3} b^{16}$

Solution 39解答 39

$\frac{1152 y^{2}}{x^{7}}$

Solution 41解答 41

$f_{2} (x) = x^{4} + 12 x^{3}$

Solution 43解答 43

$f_{4} (x) = x^{4} + 12 x^{3} + 54 x^{2} + 108 x$

Solution 45解答 45

$590, 625 x^{5} y^{2}$

Solution 47解答 47

$k - 1$

Solution 49解答 49

The expression ${(x^{3} + 2 y^{2} - z)}^{5}$ cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.

式 ${(x^{3} + 2 y^{2} - z)}^{5}$ は二項式に書き直せないので、二項定理では展開できない。

13.7 Section Exercises13.7 節末問題

Solution 1解答 1

probability; The probability of an event is restricted to values between $0$ and $1,$ inclusive of $0$ and $1.$

確率。事象の確率は $0$ と $1$ のあいだ( $0$ と $1.$ を含む)の値に限られる。

Solution 3解答 3

An experiment is an activity with an observable result.

試行とは、観測できる結果を持つ行いである。

Solution 5解答 5

The probability of the union of two events occurring is a number that describes the likelihood that at least one of the events from a probability model occurs. In both a union of sets $A and B$ and a union of events $A and B,$ the union includes either $A or B$ or both. The difference is that a union of sets results in another set, while the union of events is a probability, so it is always a numerical value between $0$ and $1.$

二つの事象の和事象が起こる確率とは、確率のモデルの事象の少なくとも一つが起こる起こりやすさを述べる数である。集合の和 $A および B$ でも事象の和 $A および B$ でも、和には $A または B$ のどちらか、または両方が含まれる。違うのは、集合の和は別の集合になるのに対し、事象の和は確率であり、つねに $0$ と $1.$ のあいだの数値になる点である。

Solution 7解答 7

$\frac{1}{2} .$

Solution 9解答 9

$\frac{5}{8} .$

Solution 11解答 11

$\frac{1}{2} .$

Solution 13解答 13

$\frac{3}{8} .$

Solution 15解答 15

$\frac{1}{4} .$

Solution 17解答 17

$\frac{3}{4} .$

Solution 19解答 19

$\frac{3}{8} .$

Solution 21解答 21

$\frac{1}{8} .$

Solution 23解答 23

$\frac{15}{16} .$

Solution 25解答 25

$\frac{5}{8} .$

Solution 27解答 27

$\frac{1}{13} .$

Solution 29解答 29

$\frac{1}{26} .$

Solution 31解答 31

$\frac{12}{13} .$

Solution 33解答 33

	1	2	3	4	5	6
1	(1,1) 2	(1,2) 3	(1,3) 4	(1,4) 5	(1,5) 6	(1,6) 7
2	(2,1) 3	(2,2) 4	(2,3) 5	(2,4) 6	(2,5) 7	(2,6) 8
3	(3,1) 4	(3,2) 5	(3,3) 6	(3,4) 7	(3,5) 8	(3,6) 9
4	(4,1) 5	(4,2) 6	(4,3) 7	(4,4) 8	(4,5) 9	(4,6) 10
5	(5,1) 6	(5,2) 7	(5,3) 8	(5,4) 9	(5,5) 10	(5,6) 11
6	(6,1) 7	(6,2) 8	(6,3) 9	(6,4) 10	(6,5) 11	(6,6) 12

Solution 35解答 35

$\frac{5}{12} .$

Solution 37解答 37

$0.$

Solution 39解答 39

$\frac{4}{9} .$

Solution 41解答 41

$\frac{1}{4} .$

Solution 43解答 43

$\frac{5}{8}$

Solution 45解答 45

$\frac{8}{13}$

Solution 47解答 47

$\frac{C (12, 5)}{C (48, 5)} = \frac{1}{2162}$

Solution 49解答 49

$\frac{C (12, 3) C (36, 2)}{C (48, 5)} = \frac{175}{2162}$

Solution 51解答 51

$\frac{C (20, 3) C (60, 17)}{C (80, 20)} \approx 12.49 %$

Solution 53解答 53

$\frac{C (20, 5) C (60, 15)}{C (80, 20)} \approx 23.33 %$

Solution 55解答 55

$20.50 + 23.33 - 12.49 = 31.34 %$

Solution 57解答 57

$\frac{C (40000000, 1) C (277000000, 4)}{C (317000000, 5)} = 36.78 %$

Solution 59解答 59

$\frac{C (40000000, 4) C (277000000, 1)}{C (317000000, 5)} = 0.11 %$

Review Exercises復習問題

Solution 1解答 1

$2, 4, 7, 11$

Solution 3解答 3

$13, 103, 1003, 10003$

Solution 5解答 5

The sequence is arithmetic. The common difference is $d = \frac{5}{3} .$

この数列は等差数列である。公差は $d = \frac{5}{3}$ である。

Solution 7解答 7

$18, 10, 2, - 6, - 14$

Solution 9解答 9

$a_{1} = - 20, a_{n} = a_{n - 1} + 10$

Solution 11解答 11

$a_{n} = \frac{1}{3} n + \frac{13}{24}$

Solution 13解答 13

$r = 2$

Solution 15解答 15

4, 16, 64, 256, 1024

Solution 17解答 17

$3, 12, 48, 192, 768$

Solution 19解答 19

$a_{n} = - \frac{1}{5} \cdot {(\frac{1}{3})}^{n - 1}$

Solution 21解答 21

$\sum_{m = 0}^{5} (\frac{1}{2} m + 5) .$

Solution 23解答 23

$S_{11} = 110$

Solution 25解答 25

$S_{9} \approx 23.95$

Solution 27解答 27

$S = \frac{135}{4}$

Solution 29解答 29

$5,617.61

Solution 31解答 31

Solution 33解答 33

$10^{4} = 10, 000$

Solution 35解答 35

$P (18, 4) = 73, 440$

Solution 37解答 37

$C (15, 6) = 5005$

Solution 39解答 39

$2^{50} = 1.13 \times 10^{15}$

Solution 41解答 41

$\frac{8!}{3! 2!} = 3360$

Solution 43解答 43

$490, 314$

Solution 45解答 45

$131, 072 a^{17} + 1, 114, 112 a^{16} b + 4, 456, 448 a^{15} b^{2}$

Solution 47解答 47

	1	2	3	4	5	6
1	1,1	1,2	1,3	1,4	1,5	1,6
2	2,1	2,2	2,3	2,4	2,5	2,6
3	3,1	3,2	3,3	3,4	3,5	3,6
4	4,1	4,2	4,3	4,4	4,5	4,6
5	5,1	5,2	5,3	5,4	5,5	5,6
6	6,1	6,2	6,3	6,4	6,5	6,6

Solution 49解答 49

$\frac{1}{6}$

Solution 51解答 51

$\frac{5}{9}$

Solution 53解答 53

$\frac{4}{9}$

Solution 55解答 55

$1 - \frac{C (350, 8)}{C (500, 8)} \approx 94.4 %$

Solution 57解答 57

$\frac{C (150, 3) C (350, 5)}{C (500, 8)} \approx 25.6 %$

Practice Test実力テスト

Solution 1解答 1

$- 14, - 6, - 2, 0$

Solution 3解答 3

The sequence is arithmetic. The common difference is $d = 0.9.$

この数列は等差数列である。公差は $d = 0.9.$ である。

Solution 5解答 5

$a_{1} = - 2, a_{n} = a_{n - 1} - \frac{3}{2}; a_{22} = - \frac{67}{2}$

Solution 7解答 7

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

この数列は等比数列である。公比は $r = \frac{1}{2}$ である。

Solution 9解答 9

$a_{1} = 1, a_{n} = - \frac{1}{2} \cdot a_{n}_{- 1}$

Solution 11解答 11

$\sum_{k = - 3}^{15} (3 k^{2} - \frac{5}{6} k)$

Solution 13解答 13

$S_{7} = - 2604.2$

Solution 15解答 15

Total in account: $$ 634, 261.20;$ Interest earned: $$ 508, 261.20$

口座の合計: $$ 634, 261.20;$ 得た利息: $$ 508, 261.20$

Solution 17解答 17

$5 \times 3 \times 2 \times 3 \times 2 = 180$

Solution 19解答 19

$C (15, 3) = 455$

Solution 21解答 21

$\frac{10!}{2! 3! 2!} = 151, 200$

Solution 23解答 23

$\frac{429 x^{14}}{16}$

Solution 25解答 25

$\frac{4}{7}$

Solution 27解答 27

$\frac{5}{7}$

Solution 29解答 29

$\frac{C (14, 3) C (26, 4)}{C (40, 7)} \approx 29.2 %$

	1	2	3	4	5	6
1	1,1	1,2	1,3	1,4	1,5	1,6
2	2,1	2,2	2,3	2,4	2,5	2,6
3	3,1	3,2	3,3	3,4	3,5	3,6
4	4,1	4,2	4,3	4,4	4,5	4,6
5	5,1	5,2	5,3	5,4	5,5	5,6
6	6,1	6,2	6,3	6,4	6,5	6,6

	1	2	3	4	5	6
1	1,1	1,2	1,3	1,4	1,5	1,6
2	2,1	2,2	2,3	2,4	2,5	2,6
3	3,1	3,2	3,3	3,4	3,5	3,6
4	4,1	4,2	4,3	4,4	4,5	4,6
5	5,1	5,2	5,3	5,4	5,5	5,6
6	6,1	6,2	6,3	6,4	6,5	6,6

	1	2	3	4	5	6
1	1,1	1,2	1,3	1,4	1,5	1,6
2	2,1	2,2	2,3	2,4	2,5	2,6
3	3,1	3,2	3,3	3,4	3,5	3,6
4	4,1	4,2	4,3	4,4	4,5	4,6
5	5,1	5,2	5,3	5,4	5,5	5,6
6	6,1	6,2	6,3	6,4	6,5	6,6