Course: 14.13 答复 -- -- 第13章:顺序和系列

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答复----第13章:序列和系列
Section 13.2 Sequences
::第13.2节顺序

Review
::回顾

39, 45, 51

-324, 972, -2,916

35, 31, 27

0.01, 0.001, 0.0001

16, 32, 64

$\begin{aligned} - \frac{1}{6}, - \frac{3}{7}, - \frac{5}{8} \end{aligned}$

9, 36

$\begin{aligned} \frac{4}{5}, \frac{6}{7} \end{aligned}$

5, 20

8, 216

6, 45

15, 35

3, 8

$\begin{aligned} a_{n} = {(- 2)}^{n - 1} \cdot \frac{1}{4} \end{aligned}$
::an=( -2)n- 1- 14

$\begin{aligned} a_{n} = 5 + 6 (n - 1) \end{aligned}$
::a=5+6(n- 1)

$\begin{aligned} a_{n} = 33 - 5 (n - 1) \end{aligned}$
::a=33-5(n-1)

$\begin{aligned} a_{n} = 4^{n - 1} \end{aligned}$
::an=4n-1

$\begin{aligned} a_{n} = 21 + 9 (n - 1) \end{aligned}$
::a=21+9(n- 1)

$\begin{aligned} a_{n} = \frac{1}{2} n^{2} + \frac{5}{2} n - 2 \end{aligned}$
::an=12n2+52n-2

$\begin{aligned} a_{n} = \frac{n}{n + 1} \end{aligned}$
::an=nn+1

$\begin{aligned} a_{n} = a_{n - 1} + a_{n - 2} \end{aligned}$
::an=an-1+an-2

9, 11, 13, 15, 17, 27

-6, -11, -16, -251

1, 3, 7, 1,023

$\begin{aligned} \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{256} \end{aligned}$

1, 3, 6, 10, 210

Explore More

$\begin{aligned} a_{n} = 9 + 29 (n - 1) \end{aligned}$
::a=9+29(n- 1)

$\begin{aligned} a_{n} = 4800 - 200 (n - 1) \end{aligned}$
::an=4800-200(n- 1)

15,30,60,120,240. Rule = $\begin{aligned} a_{n} = 7.5 {(2)}^{n} \end{aligned}$
::15,30,60,120,240. 规则==7.5(2)n

$\begin{aligned} a_{n} = \frac{3 n - 2}{3 n} \end{aligned}$
::an=3n-23n

Section 13.3: Arithmetic Sequences
::第13.3节:自学序列

Review

$\begin{aligned} a_{n} = n + 1 \end{aligned}$
::an=n+1 =n+1

Not an arithmetic sequence
::不是算术序列

$\begin{aligned} a_{n} = 5 - 5 (n - 1) \end{aligned}$
::a=5-5(n-1)

Not an arithmetic sequence
::不是算术序列

$\begin{aligned} a_{n} = 3 (n - 1) \end{aligned}$
::an=3(n- 1)

$\begin{aligned} a_{n} = - n + 14 \end{aligned}$
::a=-n+14 =-n+14

Not an arithmetic sequence
::不是算术序列

$\begin{aligned} a_{n} = a + 2 (n - 1) \end{aligned}$
::a=a+2(n-1)

$\begin{aligned} a_{n} = 15 - 8 (n - 1) \end{aligned}$
::a=15-8(n-1)

$\begin{aligned} a_{n} = - 10 + \frac{1}{2} (n - 1) \end{aligned}$
::a=-10+12(n-1)

$\begin{aligned} a_{n} = 28 - 2 (n - 1) \end{aligned}$
::a=28-2(n-1)

$\begin{aligned} a_{n} = - 15 + 3 (n - 1) \end{aligned}$
::a= 15+3(n- 1)

$\begin{aligned} a_{n} = 84 - 11 (n - 1) \end{aligned}$
::an=84-11(n-1)

$\begin{aligned} a_{n} = - 10 + 7 (n - 1) \end{aligned}$
::a=-10+7(n-1)

$\begin{aligned} a_{n} = 2 + 3 (n - 2) \end{aligned}$
::an=2+3(n-2)

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$\begin{aligned} a_{n} = 19, 500 - 500 (n - 1) \end{aligned}$
::a=19 500-500(n-1)

a. $\begin{aligned} a_{n} = \frac{1}{1 + 3 (n - 1)} . \end{aligned}$ b. $\begin{aligned} a_{n} = \frac{1}{\frac{1}{2} + \frac{1}{6} (n - 1)} . \end{aligned}$ Both are harmonic.
::a. a=11+3(n-1),b. a=112+16(n-1),两者均为口音。

$\begin{aligned} 1, \frac{5}{4}, \frac{3}{2} \end{aligned}$

No
::否无

Yes, $\begin{aligned} n = 4 \end{aligned}$
::是, n=4

Section 13.4 Geometric Sequence
::第13.4节几何序列

Review

Arithmetic
::测量学

Geometric
::几何

Neither
::中无

Geometric
::几何

Arithmetic
::测量学

Neither
::中无

$\begin{aligned} a_{n} = 32 {(\frac{3}{2})}^{n - 1}; \end{aligned}$ 32, 48, 72, 108, 162
::=32(32)n-1;32,48,72,108,162

$\begin{aligned} a_{n} = - 81 {(\frac{- 1}{3})}^{n - 1}; \end{aligned}$ -81, 27, -9 ,3, -1
::a=-81(- 13- 13n-1);- 81, 27, 9, 3, - 1

$\begin{aligned} a_{n} = 7 {(2)}^{n - 1}; \end{aligned}$ ;7, 14, 28, 56, 112
::an=7(2)n-1; ; 7, 14, 28, 56, 112

$\begin{aligned} a_{n} = \frac{8}{125} {(\frac{- 5}{2})}^{n - 1}; \end{aligned}$    $\begin{aligned} \frac{8}{125}, \frac{- 4}{25}, \frac{2}{5}, 1, \frac{5}{2} \end{aligned}$
::an=8125(-52)n-1;8125,-425,25,1,52

$\begin{aligned} a_{n} = 162 {(\frac{2}{3})}^{n - 1} \end{aligned}$
::=162(23)n-1

$\begin{aligned} a_{n} = - 625 {(\frac{3}{5})}^{n - 1} \end{aligned}$
::an=-625( 35n- 1)

$\begin{aligned} a_{n} = \frac{9}{4} {(\frac{- 2}{3})}^{n - 1} \end{aligned}$
::an=94(- 23- 23n- 1)

$\begin{aligned} a_{n} = 3 {(5)}^{n - 1} \end{aligned}$
::an=3(5)n-1

$\begin{aligned} a_{n} = 5 {(2)}^{n - 1} \end{aligned}$
::an=5(2)n-1

$\begin{aligned} a_{n} = \frac{1}{2} {(- 4)}^{n - 1} \end{aligned}$
::an=12( 4- en- 1)

$\begin{aligned} a_{n} = 54 {(\frac{2}{3})}^{n - 1} \end{aligned}$
::an=54( 23n- 1)

$\begin{aligned} a_{n} = 160 {(\frac{- 3}{4})}^{n - 1} \end{aligned}$
::an=160(- 34n- 1)

$\begin{aligned} a_{n} = 1.732 {(\frac{6}{5})}^{n - 1} \end{aligned}$
::a= 1.732(65-1)

$\begin{aligned} a_{n} = 320 {(\frac{- 1}{2})}^{n - 1} \end{aligned}$
::a=320(-12)n-1

$\begin{aligned} a_{n} = 0.857 {(\frac{6}{7})}^{n - 1} \end{aligned}$
::an=0.857(67n-1)

$\begin{aligned} a_{n} = \frac{11}{8} {(2)}^{n - 1} \end{aligned}$
::an=118(2)n-1

$\begin{aligned} a_{n} = 24 {(\frac{3}{2})}^{n - 1} \end{aligned}$
::an=24 (32)n- 1

$\begin{aligned} a_{n} = \frac{343}{216} {(\frac{6}{7})}^{n - 1} \end{aligned}$
::an=343216(67n-1)

$\begin{aligned} a_{n} = 2 {(3)}^{n - 1} \end{aligned}$
::an=2(3)n-1

$\begin{aligned} a_{n} = 128 {(\frac{3}{2})}^{n - 1} \end{aligned}$
::an=128(32)n-1

$\begin{aligned} a_{n} = \frac{8}{27} {(\frac{3}{2})}^{n - 1} \end{aligned}$
::an=827(32)n-1

$\begin{aligned} a_{n} = \frac{1}{12} {(- 2)}^{n - 1} \end{aligned}$
::an=112 (-2- en- 1)

Explore More

$ 103,946.41

$29,646.88

$\begin{aligned} a_{n} = 80 {(\frac{9}{10})}^{n - 1} \end{aligned}$
::an=80( 9910,000n- 1)

$\begin{aligned} a_{n} = 4 e^{0.693 n} \end{aligned}$
::an=4e0.693n

$34,000

$93,000

Section 13.5 Series and Summation Notation
::第13.5节系列和总和编号

Review

2, 4, 6, 8, 10; 30

8, 9, 10, 11; 38

70, 88, 108, 130, 154, 180; 730

3, 6, 10, 15, 21; 55

4, 5, 7, 11, 19, 35; 81

8, 8.5, 9, 9.5, 10, 10.5; 55.5

-25, -24, -23, -22, -21, ... , 21, 22, 23, 24, 25; 0

16, 8, 4, 2, 1; 31

27.5, 39, 52.5, 68, 85.5, 105, 126.5, 150; 654

0.5, 1, 1.5, 2, 2.5, ... , 49, 49.5, 50; 2,525

1, 2, 3, 4, 5, ... , 198, 199, 200; 20,100

a. 5, 7, 9, 11, 13; 45.
b. 15+[2, 4, 6, 8, 10]; 45
::a. 5、7、9、11、13;45 b. 15+[2、4、6、8、10];45

a. 1, 3, 6, 10, 15; 35
b. 0.5[2, 6, 12, 20, 30]; 35
::a. 1,3,6,10,15;35b.0.5[2,6,12,20,30];35

a. 4, 32, 108, 256, 500; 900.
b. 4[1, 8, 27, 64, 125]; 900
::a. 4、32、108、256、500;900b. 4[1、8、27、64、125];900;

Explore More

140 oranges
::140 橙色

64 hexagons
::64 六边形

$13,335

4,200 tourists
::4 200名游客

3,150 pairs of shoes
::3 150双鞋子

51 deer
::51鹿

Section 13.6 Finding the Sum of a Finite Arithmetic Series
::第13.6节确定一个有限自算系列的总和

Review

1,469

-84

2,499

13

875

861

240

9,900

91

-84

361

180

1,207

-483

63

Explore More

648

$1,378

$50,400

$14,400

$12,500

Section 13.7 Finding the Sum of Finite Geometric Series
::第13.7节查找有限几何系列总和

Review

$\begin{aligned} \frac{844}{9} \end{aligned}$

99

5

$\begin{aligned} \frac{1, 031}{5} \end{aligned}$

$\begin{aligned} \frac{181}{384} \end{aligned}$

$\begin{aligned} \frac{- 15}{8} \end{aligned}$

$\begin{aligned} \frac{5, 461}{128} \end{aligned}$

$\begin{aligned} \frac{11, 529}{2, 000} \end{aligned}$

$\begin{aligned} \frac{43}{2} \end{aligned}$

$\begin{aligned} a_{n} = 3 {(- 2)}^{n - 1} \end{aligned}$
::an= 3 (- 2- 个n- 1)

$\begin{aligned} a_{n} = 216 {(\frac{5}{6})}^{n - 1} \end{aligned}$
::an= 216(56)n- 1

$\begin{aligned} a_{n} = 2 {(- 3)}^{n - 1} \end{aligned}$
::an= (2)- (-)n- 1

$\begin{aligned} a_{n} = 96 {(\frac{- 1}{2})}^{n - 1} \end{aligned}$
::an=96(-12)n- 1

Explore More

$23,518.36

$1,488.78

$918.21

2,782

94 billion
::940亿

2,136.8 kg
::2,136.8公斤

36,172 views
::36 172 意见

$2,768.42

Section 13.8 Partial Sums and Finding the Sum of an Infinite Geometric Series
::第13.8节部分总和和和无限几何系列的计算总和

Review

5, 7.5, 8.75, 9.375, 9.6875; converges
::5、7.5、8.75、9.375、9.6875;交汇

2, 3.5, 4.625, 5.46875, 6.1015625; converges
::2, 3.5, 4.625, 5.46875, 6.1015625; 交汇

10, 19, 27.1, 34.39, 40.951; converges
::10、19、27.1、34.39、40.951;交汇

8, 16.24, 24.7272, 33.469016, 42.47308648; diverges
::8、16.24、24.772、33.469016、42.47308648;差异

0.5, 1.5, 3, 5, 7.5; diverges
::0.5、1.5、3、5、7.5;差异

10, 15, 18.3333, 20.8333, 22.8333; diverges
::10、15、18.3333、20.83333、22.833;差异

0.5, 0.875, 1.15625, 1.3671875, 1.525390625; converges
::0.5、0.875、1.15625、1.3671875、1.5255390625;交汇点

$\begin{aligned} 1, \frac{5}{4}, \frac{49}{36}, \frac{205}{144}, \frac{5, 269}{3, 600}; \end{aligned}$ converges
::1 54 4936 205144 5 2693 600;汇合

60, 66, 66.6, 66.66, 66.666; converges
::60、66、66、66.6、6666、66.666;交汇

5.01, 10.03, 15.06, 20.1, 25.15;  diverges
::5.01、10.03、15.06、20.1、25.15;差异

2, 3.75, 5.28125, 6.62109375, 7.793457031; converges
::2,3.75,5.28125,6.62109375,7.79457031;交汇

15

Not possible
::办办办办办办办办办办办办办办

1.5

Not possible
::办办办办办办办办办办办办办办

10

$\begin{aligned} \frac{7}{8} \end{aligned}$

2

Not possible
::办办办办办办办办办办办办办办

4

180

Not possible
::办办办办办办办办办办办办办办

125

Not possible
::办办办办办办办办办办办办办办

$\begin{aligned} 16 \frac{2}{3} \approx 16.6667 \end{aligned}$

$\begin{aligned} - \frac{48}{7} or - 6 \frac{6}{7} \approx - 6.85714 \end{aligned}$
::-487或-667-6.85714

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a. Numbers 5 and 10 are arithmetic. None have a finite sum as they are divergent. b. All but Numbers 5 and 10 are geometric. Convergent series have finite sums. c. A convergent series has a limit that exists, whereas a divergent series does not. d. Answers will vary.
::a. 数字5和10是算术。没有一个数字因差异而有一定的数值。b. 除数字5和10外,所有数字5和10均为几何数。趋同序列有一定的数值。c. 集合序列有一个存在的限制,而不同序列则没有。d. 答复将有所不同。

$\begin{aligned} 1, \frac{1}{3}, \frac{1}{6}, \frac{1}{10}, \frac{1}{15}; \end{aligned}$ the series is convergent.
::1,13,16,11,110,115;该系列是聚合的。

$\begin{aligned} 1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \frac{1}{25}; \end{aligned}$ the series is convergent.
::1,14,19,116,125; 该序列是集合的。

Section 13.9 Pascal's Triangle and Coefficients in the Expansion of Binomials
::第13.9节帕斯卡尔的三角和二民族扩张中的系数

Review

1      6      15     20     15     6     1

1     12     66     220     495     792      924     792     495     220     66     12      1

3

7

15

1

84

84

35

2,380

54,627,300

1

Explore More

The 2nd diagonal is the row number minus 1.
::第二个对角线是行号减1。

The sum of each row of Pascal's Triangle is equal to the powers of 2. Therefore, $\begin{aligned} s u m = 2^{n} . \end{aligned}$
::帕斯卡尔三角形每行的总和等于 2 的功率, 因此, sum=2n 。

The 1st 10 triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, and 55.
::第10个三角形数字为1、3、6、10、15、21、28、36、45和55。

The sum of the 1st n rows of Pascal's Triangle is equal to the n ^th Mersenne number, M _n . So when $\begin{aligned} n = 5, \end{aligned}$    $\begin{aligned} M_{5} = 2^{5} - 1 = 31, \end{aligned}$ which is the sum of the numbers in rows 1 to 5. The same is true when $\begin{aligned} n = 6. \end{aligned}$    $\begin{aligned} M_{6} = 2^{6} - 1 = 63 \end{aligned}$ is the sum of the 1st 6 rows in Pascal's Triangle.
::Pascal三角区第一 n 行的总和等于 nth Mersenne 编号Mn。当 n= 5, M5= 25-1=31时, 这是第 1 至 5. 行数字的总和。当 n= 6. M6= 26 - 1= 63 是 Pascal三角区第 1 6 行的总和时, 情况也是如此。

Section 13.10 The Binomial Theorem
::第13.10节

Review

$\begin{aligned} - a^{7} + 7 a^{6} x - 21 a^{5} x^{2} + 35 a^{4} x^{3} - 35 a^{3} x^{4} + 21 a^{2} x^{5} - 7 a x^{6} + x^{7} \end{aligned}$
::-7+7a6x-21a5x2+35a4x3-35a3x4+21a2x5-7a6+7x7

$\begin{aligned} 16 a^{4} + 96 a^{3} + 216 a^{2} + 216 a + 81 \end{aligned}$
::16a4+96a3+216a2+216a+81a+81

$\begin{aligned} - 3, 920 x^{2} \end{aligned}$
::- 3,920x2

$\begin{aligned} 40, 824 x^{5} \end{aligned}$
::40,824x5

$\begin{aligned} 1, 312, 500 a^{3} \end{aligned}$
::1 312 500a3

$\begin{aligned} 4, 320 x^{3} y^{3} \end{aligned}$
::4,320x3y3

$\begin{aligned} 20, 412 x^{5} \end{aligned}$
::20,412x5

$\begin{aligned} 700 y^{6} \end{aligned}$
::700y6

$\begin{aligned} - 960 a^{3} b^{7} \end{aligned}$
::-960a3b7 -960a3b7

$\begin{aligned} 3, 240 x^{4} \end{aligned}$
::3 240x4

1,215

700

Section 13.11 Connections: The Fibonacci Sequence
::第13.11节连接:Fibonacci序列

21, 34, 55, 89, 144

The Fibonacci Sequence is an arithmetic series.
::Fibonacci序列是一个计算序列。

$\begin{aligned} X_{n} = X_{n - 1} + X_{n - 2} \end{aligned}$
::Xn=Xn-1+Xn-2

143

No. The Fibonacci Sequence diverges to infinity.
::否。 Fibonacci 序列与无穷相区别。

$\begin{aligned} F_{7} = [\begin{matrix} 6 \\ 0 \end{matrix}] + [\begin{matrix} 5 \\ 1 \end{matrix}] + [\begin{matrix} 4 \\ 2 \end{matrix}] + [\begin{matrix} 3 \\ 3 \end{matrix}] = 13 \end{aligned}$
::F7=[60]+[51]+[42]+[33]=13

Section 13.12 Connections: Zeno's Paradoxes
::第13.12节连接:Zeno的悖论

$\begin{aligned} 64, 32, 16, 8, 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8} \end{aligned}$

$\begin{aligned} a_{n} = 64 {(\frac{1}{2})}^{n - 1} \end{aligned}$
::an=64( 12)n- 1

$\begin{aligned} 64, 96, 112, 120, 124 \end{aligned}$

128

Answers will vary.
::答案将各有不同。

Section outline

General

答复----第13章:序列和系列

Section 13.2 Sequences ::第13.2节 顺序

Section 13.3: Arithmetic Sequences ::第13.3节:自学序列

Section 13.4 Geometric Sequence ::第13.4节 几何序列

Section 13.5 Series and Summation Notation ::第13.5节 系列和总和编号

Section 13.6 Finding the Sum of a Finite Arithmetic Series ::第13.6节 确定一个有限自算系列的总和

Section 13.7 Finding the Sum of Finite Geometric Series ::第13.7节 查找有限几何系列总和

Section 13.8 Partial Sums and Finding the Sum of an Infinite Geometric Series ::第13.8节 部分总和和和无限几何系列的计算总和

Section 13.9 Pascal's Triangle and Coefficients in the Expansion of Binomials ::第13.9节 帕斯卡尔的三角和二民族扩张中的系数

Section 13.10 The Binomial Theorem ::第13.10节

Section 13.11 Connections: The Fibonacci Sequence ::第13.11节连接:Fibonacci序列

Section 13.12 Connections: Zeno's Paradoxes ::第13.12节连接:Zeno的悖论

Section 13.2 Sequences
::第13.2节顺序

Section 13.3: Arithmetic Sequences
::第13.3节:自学序列

Section 13.4 Geometric Sequence
::第13.4节几何序列

Section 13.5 Series and Summation Notation
::第13.5节系列和总和编号

Section 13.6 Finding the Sum of a Finite Arithmetic Series
::第13.6节确定一个有限自算系列的总和

Section 13.7 Finding the Sum of Finite Geometric Series
::第13.7节查找有限几何系列总和

Section 13.8 Partial Sums and Finding the Sum of an Infinite Geometric Series
::第13.8节部分总和和和无限几何系列的计算总和

Section 13.9 Pascal's Triangle and Coefficients in the Expansion of Binomials
::第13.9节帕斯卡尔的三角和二民族扩张中的系数

Section 13.10 The Binomial Theorem
::第13.10节

Section 13.11 Connections: The Fibonacci Sequence
::第13.11节连接:Fibonacci序列

Section 13.12 Connections: Zeno's Paradoxes
::第13.12节连接:Zeno的悖论