10.1 Define and Use Sequences and Series€¦ · Define and Use Sequences and Series Goal p Recognize and write rules for number patterns. VOCABULARY Sequence A function whose domain

Your Notes

Define and Use Sequences and SeriesGoal p Recognize and write rules for number patterns.

VOCABULARY

Sequence

Terms

Series

Summation notation

Sigma notation

SEQUENCES

A sequence is a function whose domain is a set of integers. If a domain is not specified, it is understood that the domain starts with 1. The values in the range are called the of the sequence.

Domain: 1 2 3 4 . . . n The relative position of each term

Range: a1 a2 a3 a4 . . . an Terms of the sequence

A sequence has a limited number of terms. An sequence continues without stopping.

Finite sequence: 2, 4, 6, 8 Infinite Sequence: 2, 4, 6, 8, . . .

A sequence can be specified by an equation, or . For example, both sequences above can be described by the rule an 5 2n or f(n) 5 2n.

Copyright © Holt McDougal. All rights reserved. Lesson 10.1 • Algebra 2 Notetaking Guide 261

10.1

Your Notes

Define and Use Sequences and SeriesGoal p Recognize and write rules for number patterns.

VOCABULARY

Sequence A function whose domain is a set of consecutive integers

Terms The values in the range of a sequence

Series The expression that results when the terms of a sequence are added together

Summation notation Notation for a series that represents the sum of the terms

Sigma notation Another name for summation notation, which uses the uppercase Greek letter, sigma, written S

SEQUENCES

A sequence is a function whose domain is a set of consecutive integers. If a domain is not specified, it is understood that the domain starts with 1. The values in the range are called the terms of the sequence.

Domain: 1 2 3 4 . . . n The relative position of each term

Range: a1 a2 a3 a4 . . . an Terms of the sequence

A finite sequence has a limited number of terms. An infinite sequence continues without stopping.

Finite sequence: 2, 4, 6, 8 Infinite Sequence: 2, 4, 6, 8, . . .

A sequence can be specified by an equation, or rule . For example, both sequences above can be described by the rule an 5 2n or f(n) 5 2n.


10.1

Your Notes

Write the first six terms of an 5 2n 1 1.

a1 5 5 1st term

a2 5 5 2nd term

a3 5 5 3rd term

a4 5 5 4th term

a5 5 5 5th term

a6 5 5 6th term

Example 1 Write terms of sequences

Describe the pattern, write the next term, and write a rule for the nth term of the sequence (a) 1, 4, 9, 16, . . . and (b) 0, 7, 26, 63, . . ..

Solution

a. You can write the terms as 2, 2, 2, 2, . . .. The next term is a5 5 5 . A rule for the nth term is an 5 .

b. You can write the terms as 2 1, 2 1, 2 1, 2 1, . . .. The next term is

a5 5 2 1 5 . A rule for the nth term is an 5 .

Example 2 Write rules for sequences

1. Write the first six terms of the sequence f (n) 5 3n 2 7.

2. For the sequence 23, 9, 227, 81, . . ., describe the pattern, write the next term, and write a rule for the nth term.

Checkpoint Complete the following exercises.

262 Lesson 10.1 • Algebra 2 Notetaking Guide Copyright © Holt McDougal. All rights reserved.

Your Notes

Write the first six terms of an 5 2n 1 1.

a1 5 21 1 1 5 4 1st term

a2 5 22 1 1 5 8 2nd term

a3 5 23 1 1 5 16 3rd term

a4 5 24 1 1 5 32 4th term

a5 5 25 1 1 5 64 5th term

a6 5 26 1 1 5 128 6th term

Example 1 Write terms of sequences

Describe the pattern, write the next term, and write a rule for the nth term of the sequence (a) 1, 4, 9, 16, . . . and (b) 0, 7, 26, 63, . . ..

Solution

a. You can write the terms as 1 2, 2 2, 3 2, 4 2, . . .. The next term is a5 5 52 5 25 . A rule for the nth term is an 5 n2 .

b. You can write the terms as 13 2 1, 23 2 1, 33 2 1, 43 2 1, . . .. The next term is a5 5 53 2 1 5 124 . A rule for the nth term is an 5 n3 2 1 .

Example 2 Write rules for sequences

1. Write the first six terms of the sequence f (n) 5 3n 2 7.

24, 21, 2, 5, 8, 11

2. For the sequence 23, 9, 227, 81, . . ., describe the pattern, write the next term, and write a rule for the nth term.

(23)1, (23)2, (23)3, (23)4; a5 5 2243; an 5 (23)n



Your NotesSERIES AND SUMMATION NOTATION

When the terms of a sequence are added together, the resulting expression is a series. A series can be finite or infinite.

Finite series: 2 1 4 1 6 1 8

Infinite series: 2 1 4 1 6 1 8 . . .

You can use notation to write a series.

2 1 4 1 6 1 8 5 i 5 1

∑ 4

2i

2 1 4 1 6 1 8 1 . . . 5 i 5 1

∑ `

2i

For both series, the index of summation is and the lower limit of summation is . The upper limit of summation is for the finite series and ( ) for the infinite series. Summation notation is also called notation because it uses the uppercase Greek letter sigma, written S.


Write the series using summation notation.

a. 4 1 7 1 10 1 . . . 1 46 b. 1 1 1 } 8 1 1 } 27 1 1 } 64 1 . . .

Solutiona. Notice that the first term is 3(1) 1 1, the second

is , the third is , and the last is . So, ai 5 where i 5 1, 2, 3, . . ., . The lower limit of summation is and the upper limit of summation is .

The summation notation for the series is .

b. Notice that for each term, the denominator is a perfect

cube. So, ai 5 where i 5 1, 2, 3, 4, . . .. The lower

limit of summation is and the upper limit of summation is .

The summation notation for the series is .

Example 3 Write series using summation notation

Your NotesSERIES AND SUMMATION NOTATION

When the terms of a sequence are added together, the resulting expression is a series. A series can be finite or infinite.

Finite series: 2 1 4 1 6 1 8

Infinite series: 2 1 4 1 6 1 8 . . .

You can use summation notation to write a series.

2 1 4 1 6 1 8 5 i 5 1

∑ 4

2i

2 1 4 1 6 1 8 1 . . . 5 i 5 1

∑ `

2i

For both series, the index of summation is i and the lower limit of summation is 1 . The upper limit of summation is 4 for the finite series and ` ( infinity ) for the infinite series. Summation notation is also called sigma notation because it uses the uppercase Greek letter sigma, written S.


Write the series using summation notation.

a. 4 1 7 1 10 1 . . . 1 46 b. 1 1 1 } 8 1 1 } 27 1 1 } 64 1 . . .

Solutiona. Notice that the first term is 3(1) 1 1, the second

is 3(2) 1 1 , the third is 3(3) 1 1 , and the last is 3(15) 1 1 . So, ai 5 3i 1 1 where i 5 1, 2, 3, . . ., 15 . The lower limit of summation is 1 and the upper limit of summation is 15 .

The summation notation for the series is i 5 1

∑ 15

3i 1 1 .

b. Notice that for each term, the denominator is a perfect

cube. So, ai 5 1 } i3 where i 5 1, 2, 3, 4, . . .. The lower

limit of summation is 1 and the upper limit of summation is infinity .

The summation notation for the series is i 5 1

∑ `

1 } i3 .

Example 3 Write series using summation notation

Homework

Your Notes

FORMULAS FOR SPECIAL SERIES

Sum of n Sum of first n Sum of squares ofterms of 1 positive integers first n positive integers

i 5 1

∑ n

1 5 n i 5 1

∑ n

i 5 n(n 1 1)

} 2 i 5 1

∑ n

i2 5 n(n 1 1)(2n 1 1)

}} 6

Find the sum of the series.

k 5 3

∑ 5

2 2 3k 5 [2 2 3( )] 1 [2 2 3( )] 1 [2 2 3( )]

5 5

Example 4 Find the sum of a series

3. 7 1 14 1 21 1 . . . 1 77

4. 24 2 8 2 12 2 16 2 . . .

Checkpoint Write the series using summation notation.

5. k 5 4

∑ 8

(k2 2 6) 6. i 5 1

∑ 28

i2

Checkpoint Find the sum of the series.

Use a formula for special series to find the sum of i 5 1

∑ 32

i.

i 5 1

∑ 32

i 5 5 5

Example 5 Use a formula for a sum


Homework

Your Notes

FORMULAS FOR SPECIAL SERIES

Sum of n Sum of first n Sum of squares ofterms of 1 positive integers first n positive integers

i 5 1

∑ n

1 5 n i 5 1

∑ n

i 5 n(n 1 1)

} 2 i 5 1

∑ n

i2 5 n(n 1 1)(2n 1 1)

}} 6

Find the sum of the series.

k 5 3

∑ 5

2 2 3k 5 [2 2 3( 3 )] 1 [2 2 3( 4 )] 1 [2 2 3( 5 )]

5 27 1 (210) 1 (213) 5 230

Example 4 Find the sum of a series

3. 7 1 14 1 21 1 . . . 1 77

i 5 1

∑ 11

7i

4. 24 2 8 2 12 2 16 2 . . .

i 5 1

∑ `

2 4i

Checkpoint Write the series using summation notation.

5. k 5 4

∑ 8

(k2 2 6) 6. i 5 1

∑ 28

i2

160 7714

Checkpoint Find the sum of the series.

Use a formula for special series to find the sum of i 5 1

∑ 32

i.

i 5 1

∑ 32

i 5 32(32 1 1)

} 2 5 32(33)

} 2 5 528

Example 5 Use a formula for a sum


Your Notes


10.2 Analyze Arithmetic Sequences and SeriesGoal p Study arithmetic sequences and series.

VOCABULARY

Arithmetic sequence

Common difference

Arithmetic series

Tell whether the sequence 25, 23, 21, 1, 3, . . . is arithmetic.

Find the differences of consecutive terms.

a2 2 a1 5 5

a3 2 a2 5 5

a4 2 a3 5 5

a5 2 a4 5 5

Each difference is , so the sequence arithmetic.

Example 1 Identify arithmetic sequences

1. 32, 27, 21, 17, 10, . . .

Checkpoint Decide whether the sequence is arithmetic.

RULE FOR AN ARITHMETIC SEQUENCE

The nth term of an arithmetic sequence with first term a1 and common difference d is given by:

an 5 a1 1 (n 2 1)d

Your Notes


10.2 Analyze Arithmetic Sequences and SeriesGoal p Study arithmetic sequences and series.

VOCABULARY

Arithmetic sequence A sequence in which the difference between consecutive terms is constant

Common difference The constant difference between terms of an arithmetic sequence, denoted by d

Arithmetic series The expression formed by adding the terms of an arithmetic sequence, denoted by Sn

Tell whether the sequence 25, 23, 21, 1, 3, . . . is arithmetic.

Find the differences of consecutive terms.

a2 2 a1 5 23 2 (25) 5 2

a3 2 a2 5 21 2 (23) 5 2

a4 2 a3 5 1 2 (21) 5 2

a5 2 a4 5 3 2 1 5 2

Each difference is 2 , so the sequence is arithmetic.

Example 1 Identify arithmetic sequences

1. 32, 27, 21, 17, 10, . . .

not arithmetic

Checkpoint Decide whether the sequence is arithmetic.

RULE FOR AN ARITHMETIC SEQUENCE

The nth term of an arithmetic sequence with first term a1 and common difference d is given by:

an 5 a1 1 (n 2 1)d

Your Notes

Write a rule for the nth term of the sequence. Then find a19.

a. 2, 9, 16, 23, . . . b. 57, 45, 33, 21, . . .

Solutiona. The sequence is arithmetic with first term a1 5 2 and

common difference d 5 5 . So, a rule for the nth term is:

an 5 a1 1 (n 2 1)d Write general rule.

5 1 (n 2 1) Substitute for a1 and d.

5 Simplify.

The 19th term is a19 5 5 .

b. The sequence is arithmetic with first term a1 5 57 and common difference d 5 5 . So, a rule for the nth term is:


5 1 (n 2 1)( ) Substitute for a1 and d.

5 Simplify.


Example 2 Write a rule for the nth term

2. 9, 5, 1, 23, . . .

3. 215, 29, 23, 3, . . .

Checkpoint Write a rule for the nth term of the arithmetic sequence. Then find a22.


Your Notes

Write a rule for the nth term of the sequence. Then find a19.

a. 2, 9, 16, 23, . . . b. 57, 45, 33, 21, . . .

Solutiona. The sequence is arithmetic with first term a1 5 2 and

common difference d 5 9 2 2 5 7 . So, a rule for the nth term is:


5 2 1 (n 2 1) 7 Substitute for a1 and d.

5 25 1 7n Simplify.

The 19th term is a19 5 25 1 7(19) 5 128 .

b. The sequence is arithmetic with first term a1 5 57 and common difference d 5 45 2 57 5 212 . So, a rule for the nth term is:


5 57 1 (n 2 1)( 212 ) Substitute for a1 and d.

5 69 2 12n Simplify.

The 19th term is a19 5 69 2 12(19) 5 2159 .


2. 9, 5, 1, 23, . . .

an 5 13 2 4n, 275

3. 215, 29, 23, 3, . . .

an 5 221 1 6n, 111



Your Notes


One term of an arithmetic sequence is a11 5 41. The common difference is d 5 5. (a) Write a rule for the nth term. (b) Graph the sequence.

a. Use the general rule to find the first term.


5 a1 1 ( 2 1) Substitute for an, n, and d.

5 a1 Solve for a1.

So, a rule for the nth term is:

an 5 1 (n 2 1) Substitute for a1 and for d.

5 Simplify.

b. Create a table of values for the

n

an

3

1

sequence. Notice that the points lie on a line.

n 1 2 3 4 5 6

an

Example 3 Write a rule given a term and common difference

Two terms of the arithmetic sequence are a6 5 7 and a22 5 87. Find a rule for the nth term.

1. Write a system of equations using an 5 a1 1 (n 2 1)d and substituting 22 for n (Equation 1) and then 6 for n (Equation 2).

a22 5 a1 1 (22 2 1)d 5 a1 1 d

a6 5 a1 1 (6 2 1)d 5 a1 1 d

2. Solve the system. 5 d

5 d

5 a1 1 ( )

5 a1

3. Find a rule for an. an 5 a1 1 (n 2 1)d

5 1 (n 2 1)

5

Example 4 Write a rule given two terms

Your Notes


One term of an arithmetic sequence is a11 5 41. The common difference is d 5 5. (a) Write a rule for the nth term. (b) Graph the sequence.



41 5 a1 1 ( 11 2 1) 5 Substitute for an, n, and d.

29 5 a1 Solve for a1.


an 5 29 1 (n 2 1) 5 Substitute for a1 and for d.

5 214 1 5n Simplify.

b. Create a table of values for the

n

an

3

1

sequence. Notice that the points lie on a line.

n 1 2 3 4 5 6

an 29 24 1 6 11 16

Example 3 Write a rule given a term and common difference

Two terms of the arithmetic sequence are a6 5 7 and a22 5 87. Find a rule for the nth term.

1. Write a system of equations using an 5 a1 1 (n 2 1)d and substituting 22 for n (Equation 1) and then 6 for n (Equation 2).

a22 5 a1 1 (22 2 1)d 87 5 a1 1 21 d

a6 5 a1 1 (6 2 1)d 7 5 a1 1 5 d

2. Solve the system. 80 5 16 d

5 5 d

87 5 a1 1 21 ( 5 )

218 5 a1

3. Find a rule for an. an 5 a1 1 (n 2 1)d

5 218 1 (n 2 1) 5

5 223 1 5n


Homework

Your Notes

4. a15 5 107, d 5 12

5. a5 5 91, a20 5 1


THE SUM OF A FINITE ARITHMETIC SERIES

The sum of the first n terms of an arithmetic series is:

Sn 5 n 1 a1 1 an } 2 2

In words, Sn is the of the terms, by .

Find the sum of the arithmetic series i 5 1

∑ 15

(9 1 3i).

a1 5 9 1 3( ) 5 Identify first term.

a15 5 9 1 3( ) 5 Identify last term.

S15 5 Write rule for S15.

5 Simplify.

Example 5 Find a sum

6. i 5 1

∑ 18

(77 2 4i)

Checkpoint Find the sum of the arithmetic series.


Homework

Your Notes

4. a15 5 107, d 5 12

an 5 273 1 12n, 191

5. a5 5 91, a20 5 1

an 5 121 2 6n, 211


THE SUM OF A FINITE ARITHMETIC SERIES

The sum of the first n terms of an arithmetic series is:

Sn 5 n 1 a1 1 an } 2 2

In words, Sn is the mean of the first and nth terms, multiplied by the number of terms .

Find the sum of the arithmetic series i 5 1

∑ 15

(9 1 3i).

a1 5 9 1 3( 1 ) 5 12 Identify first term.

a15 5 9 1 3( 15 ) 5 54 Identify last term.

S15 5 15 1 12 1 54 } 2 2 Write rule for S15.

5 495 Simplify.

Example 5 Find a sum

6. i 5 1

∑ 18

(77 2 4i)

702

Checkpoint Find the sum of the arithmetic series.


Your Notes


10.3 Analyze Geometric Sequences and SeriesGoal p Study geometric sequences and series.

VOCABULARY

Geometric sequence

Common ratio

Geometric series

Tell whether the sequence 1, 24, 16, 264, 256, . . . is geometric.

To decide whether a sequence is geometric, find the ratios of consecutive terms.

a2

} a1 5 24

} 1 5 24 a3

} a2 5 5

a4

} a3 5 5

a5 } a4 5 5

Each ratio is , so the sequence geometric.

Example 1 Identify geometric sequences.

1. 512, 128, 64, 8, . . .

Checkpoint Tell whether the sequence is geometric.

RULE FOR A GEOMETRIC SEQUENCE

The nth term of a geometric sequence with first term a1 and common ratio r is given by: an 5 a1r n 2 1

Your Notes


10.3 Analyze Geometric Sequences and SeriesGoal p Study geometric sequences and series.

VOCABULARY

Geometric sequence A sequence in which the ratio of any term to the previous term is constant

Common ratio The constant ratio between consecutive terms of a geometric sequence, denoted by r

Geometric series The expression formed by adding the terms of a geometric sequence

Tell whether the sequence 1, 24, 16, 264, 256, . . . is geometric.

To decide whether a sequence is geometric, find the ratios of consecutive terms.

a2

} a1 5 24

} 1 5 24 a3

} a2 5 16

} 24 5 24

a4

} a3 5 264

} 16 5 24 a5

} a4 5 256

} 264 5 24

Each ratio is 24 , so the sequence is geometric.

Example 1 Identify geometric sequences.

1. 512, 128, 64, 8, . . .

not geometric

Checkpoint Tell whether the sequence is geometric.

RULE FOR A GEOMETRIC SEQUENCE

The nth term of a geometric sequence with first term a1 and common ratio r is given by: an 5 a1r n 2 1

Your Notes

Write a rule for the nth term of the sequence 972, 2324, 108, 236, . . .. Then find a10.

Solution

The sequence is geometric with first term a1 5

and common ratio r 5 5 . So, a rule for the

nth term is:

an 5 a1rn 2 1 Write general rule.

5 1 2 n 2 1 Substitute for a1 and r.



One term of a geometric sequence is a3 5 218. The common ratio is r 5 3. (a) Write a rule for the nth term. (b) Graph the sequence.



5 a1( ) 2 1 Substitute for an, r, and n.

5 a1 Solve for a1.



5 Substitute for a1 and r.

b. Create a table of values for the n

an

2201sequence. Notice that the points

lie on an exponential curve.

n 1 2 3

an

n 4 5

an

Example 3 Write a rule given a term and common ratio


Your Notes

Write a rule for the nth term of the sequence 972, 2324, 108, 236, . . .. Then find a10.

Solution

The sequence is geometric with first term a1 5 972

and common ratio r 5 2324 } 972 5 2

1 } 3 . So, a rule for the

nth term is:


5 972 1 2 1 } 3 2 n 2 1

Substitute for a1 and r.

The 10th term is a10 5 972 1 2 1 } 3 2 10 2 1

5 2 4 } 81

.


One term of a geometric sequence is a3 5 218. The common ratio is r 5 3. (a) Write a rule for the nth term. (b) Graph the sequence.



218 5 a1( 3 ) 3 2 1 Substitute for an, r, and n.




5 22(3)n 2 1 Substitute for a1 and r.

b. Create a table of values for the n

an

2201sequence. Notice that the points

lie on an exponential curve.

n 1 2 3

an 22 26 218

n 4 5

an 254 2162

Example 3 Write a rule given a term and common ratio


Your Notes


2. 14, 28, 56, 112, . . .

3. a5 5 324, r 5 23

Checkpoint Write a rule for the nth term of the geometric sequence. Then find a9.

Two terms of a geometric sequence are a2 5 10 and a7 5 2320. Find a rule for the nth term.

1. Write a system of equations using an 5 a1rn 2 1 and substituting 2 for n (Equation 1) and then 7 for n (Equation 2).

a2 5 a1r2 2 1 Equation 1

a7 5 a1r7 2 1 Equation 2

2. Solve the system. 5 a1 Solve Equation 1 for a1.

2320 5 (r6) Substitute for a1 in Equation 2.

2320 5 Simplify.

5 r Solve for r.

10 5 a1( ) Substitute in Equation 1

5 a1 Solve for a1.

3. Find a rule for an. an 5 a1rn 2 1 Write general rule.

an 5 Substitute.


Your Notes


2. 14, 28, 56, 112, . . .

an 5 14(2)n 2 1; 3584

3. a5 5 324, r 5 23

an 5 4(23)n 2 1; 26,244


Two terms of a geometric sequence are a2 5 10 and a7 5 2320. Find a rule for the nth term.

1. Write a system of equations using an 5 a1rn 2 1 and substituting 2 for n (Equation 1) and then 7 for n (Equation 2).

a2 5 a1r2 2 1 10 5 a1r Equation 1

a7 5 a1r7 2 1 2320 5 a1r6 Equation 2

2. Solve the system. 10 } r 5 a1 Solve Equation

1 for a1.

2320 5 10 } r (r6) Substitute for a1

in Equation 2.

2320 5 10r5 Simplify.

22 5 r Solve for r.

10 5 a1( 22 ) Substitute in Equation 1


3. Find a rule for an. an 5 a1rn 2 1 Write general rule.

an 5 25(22)n 2 1 Substitute.


Homework

Your Notes

4. a3 5 224, a6 5 28


THE SUM OF A FINITE GEOMETRIC SERIES

The sum of the first n terms of a geometric series with common ratio r Þ 1 is:

Sn 5 a1 1 1 2 rn } 1 2 r 2

Find the sum of the geometric series i 5 1

∑ 13

3(4)i 2 1.

a1 5 5 Identify first term.

r 5 Identify common ratio.

S13 5 a1 1 1 2 r13 } 1 2 r 2 Write rule for S13.

5 5 Substitute and simplify.

Example 5 Find the sum of a geometric series

5. i 5 1

∑ 11

7(25)n 2 1

Checkpoint Find the sum of the geometric series.


Homework

Your Notes

4. a3 5 224, a6 5 28

an 5 896 1 1 } 2 2 n 2 1

, 7 } 2


THE SUM OF A FINITE GEOMETRIC SERIES

The sum of the first n terms of a geometric series with common ratio r Þ 1 is:

Sn 5 a1 1 1 2 rn } 1 2 r 2

Find the sum of the geometric series i 5 1

∑ 13

3(4)i 2 1.

a1 5 3(4)1 2 1 5 3 Identify first term.

r 5 4 Identify common ratio.

S13 5 a1 1 1 2 r13 } 1 2 r 2 Write rule for S13.

5 3 1 1 2 413 }

1 2 4 2 5 67,108,863 Substitute and simplify.

Example 5 Find the sum of a geometric series

5. i 5 1

∑ 11

7(25)n 2 1

56,966,147

Checkpoint Find the sum of the geometric series.


Your Notes


10.4 Find Sums of Infinite Geometric SeriesGoal p Find the sums of infinite geometric series.

VOCABULARY

Partial sum

THE SUM OF AN INFINITE GEOMETRIC SERIES

The sum of an infinite geometric series with first term a1 and common ratio r is given by

S 5 a1 } 1 2 r

provided ⏐r⏐ < 1. If ⏐r⏐ ≥ 1, the series has .

Find the sum of the infinite geometric series.

a. i 5 1

∑ `

6(0.6) i 2 1 b. 1 2 2 } 3 1 4 } 9 2 8 } 27 1 . . .

c. 1 2 2 1 4 2 8 1 . . .

Solution

a. For this series, a1 5 and r 5 .

S 5 a1 } 1 2 r 5 5

b. For this series, a1 5 and r 5 .

S 5 a1 } 1 2 r 5 5

c. You know that a1 5 and a2 5 . So,

r 5 5 . Because ⏐ ⏐ 1, the sum

.

Example 1 Find sums of infinite geometric series

Your Notes


10.4 Find Sums of Infinite Geometric SeriesGoal p Find the sums of infinite geometric series.

VOCABULARY

Partial sum The sum Sn of the first n terms of an infinite series

THE SUM OF AN INFINITE GEOMETRIC SERIES

The sum of an infinite geometric series with first term a1 and common ratio r is given by

S 5 a1 } 1 2 r

provided ⏐r⏐ < 1. If ⏐r⏐ ≥ 1, the series has no sum .

Find the sum of the infinite geometric series.

a. i 5 1

∑ `

6(0.6) i 2 1 b. 1 2 2 } 3 1 4 } 9 2 8 } 27 1 . . .

c. 1 2 2 1 4 2 8 1 . . .

Solution

a. For this series, a1 5 6 and r 5 0.6 .

S 5 a1 } 1 2 r 5 6

} 1 2 0.6 5 15

b. For this series, a1 5 1 and r 5 2 2 } 3 .

S 5 a1 } 1 2 r 5 1

} 1 2 1 2 2 }

3 2 5 3 } 5

c. You know that a1 5 1 and a2 5 22 . So,

r 5 22 } 1 5 22 . Because ⏐ 22 ⏐ ≥ 1, the sum

does not exist .

Example 1 Find sums of infinite geometric series

1. k 5 1

∑ `

5 1 9 } 7 2 k 2 1

2. n 5 1

∑ `

9 1 5 } 6 2 n 2 1

3. 6 1 10 } 3 1 50

} 27 1 250 } 243 1 . . .

Checkpoint Find the sum of the infinite geometric series, if it exists.

Swings A person is given one push on a swing. On the first swing, the person travels a distance of 4 feet. On each successive swing, the person travels 75% of the distance of the previous swing. What is the total distance the person swings?

SolutionThe total distance traveled by the person is:

d 5 4 1 4( ) 1 4( )2 1 4( )3 1 . . .

5 a1 } 1 2 r Write formula for sum.


5 Simplify.

The swing travels a total distance of feet.

Example 2 Use an infinite series as a model


Your Notes

1. k 5 1

∑ `

5 1 9 } 7 2 k 2 1

does not exist

2. n 5 1

∑ `

9 1 5 } 6 2 n 2 1

54

3. 6 1 10 } 3 1 50

} 27 1 250 } 243 1 . . .

13.5

Checkpoint Find the sum of the infinite geometric series, if it exists.

Swings A person is given one push on a swing. On the first swing, the person travels a distance of 4 feet. On each successive swing, the person travels 75% of the distance of the previous swing. What is the total distance the person swings?

SolutionThe total distance traveled by the person is:

d 5 4 1 4( 0.75 ) 1 4( 0.75 )2 1 4( 0.75 )3 1 . . .


5 4 } 1 2 0.75 Substitute for a1 and r.

5 16 Simplify.

The swing travels a total distance of 16 feet.

Example 2 Use an infinite series as a model


Your Notes

Homework

Your Notes


4. In Example 2, suppose the person travels 3 feet on the first swing. What is the total distance the person swings?

Checkpoint Complete the following exercise.

Write 0.474747 . . . as a fraction in lowest terms.

Solution0.474747 . . .

5 47( ) 1 47( )2 1 47( )3 1 . . .



5 Simplify.

5 Write as a quotient of integers.

The repeating decimal 0.474747 . . . is as a fraction.

Example 3 Write a repeating decimal as a fraction

5. 0.888 . . . 6. 0.636363 . . .

Checkpoint Write the repeating decimal as a fraction.

Homework

Your Notes


4. In Example 2, suppose the person travels 3 feet on the first swing. What is the total distance the person swings?

12 feet


Write 0.474747 . . . as a fraction in lowest terms.

Solution0.474747 . . .

5 47( 0.01 ) 1 47( 0.01 )2 1 47( 0.01)3 1 . . .


5 47(0.01)

} 1 2 0.01 Substitute for a1 and r.

5 0.47 } 0.99 Simplify.

5 47 } 99 Write as a quotient of integers.

The repeating decimal 0.474747 . . . is 47 } 99 as a

fraction.

Example 3 Write a repeating decimal as a fraction

5. 0.888 . . . 6. 0.636363 . . .

8 } 9 7 } 11

Checkpoint Write the repeating decimal as a fraction.

Apply the Counting Principle and PermutationsGoal p Use the fundamental counting principle and find

permutations.

VOCABULARY

Permutation

Factorial

FUNDAMENTAL COUNTING PRINCIPLE

Two Events If one event can occur in m ways and another event can occur in n ways, then the number of ways that both events can occur is .

Three or More Events The fundamental counting principle can be extended to three or more events. For example, if three events occur in m, n, and p ways, then the number of ways that all three events can occur is

.

Pizza You are buying a pizza. You have a choice of 3 crusts, 4 cheeses, 5 meat toppings, and 8 vegetable toppings. How many different pizzas with one crust, one cheese, one meat, and one vegetable can you choose?

SolutionUse the fundamental counting principle to find the total number of pizzas. Multiply the number of crusts ( ), the number of cheeses ( ), the number of meats ( ), and the number of vegetables ( ).

Number of pizzas 5 5

Example 1 Use the fundamental counting principle

10.5


Your Notes

Apply the Counting Principle and PermutationsGoal p Use the fundamental counting principle and find

permutations.

VOCABULARY

Permutation A permutation is an ordering of n objects.

Factorial Represented by the symbol !, n factorial is defined as: n! 5 n p (n 2 1) p (n 2 2) p . . . p 3 p 2 p 1.

FUNDAMENTAL COUNTING PRINCIPLE

Two Events If one event can occur in m ways and another event can occur in n ways, then the number of ways that both events can occur is m p n .

Three or More Events The fundamental counting principle can be extended to three or more events. For example, if three events occur in m, n, and p ways, then the number of ways that all three events can occur is m p n p p .

Pizza You are buying a pizza. You have a choice of 3 crusts, 4 cheeses, 5 meat toppings, and 8 vegetable toppings. How many different pizzas with one crust, one cheese, one meat, and one vegetable can you choose?

SolutionUse the fundamental counting principle to find the total number of pizzas. Multiply the number of crusts ( 3 ), the number of cheeses ( 4 ), the number of meats ( 5 ), and the number of vegetables ( 8 ).

Number of pizzas 5 3 p 4 p 5 p 8 5 480

Example 1 Use the fundamental counting principle

10.5


Your Notes

Your Notes

Telephone Numbers A town has telephone numbers that all begin with 329 followed by four digits. How many different phone numbers are possible (a) if numbers can be repeated and (b) if numbers cannot be repeated?

a. There are choices for each digit. Use the fundamental counting principle to find the total amount of phone numbers.

Phone numbers 5 5

b. If you cannot repeat digits, there are still choices for the first number, but then only remaining choices for the second digit, choices for the third digit, and choices for the fourth digit. Use the fundamental counting principle.

Phone numbers 5 5

Example 2 Use the counting principle with repetition

1. If the pizza crust was not a choice in Example 1, how many different pizzas could be made?


Playoffs Eight teams are competing in a baseball playoff.

a. In how many different ways can the baseball teams finish the competition?

b. In how many different ways can 3 of the baseball teams finish first, second, and third?

Solutiona. There are 8! different ways that the teams can finish.

8! 5 5

b. Any of the teams can finish first, then any of the remaining teams can finish second, and then any of the remaining teams can finish third.

5

Example 3 Find the number of permutations


Your Notes

Telephone Numbers A town has telephone numbers that all begin with 329 followed by four digits. How many different phone numbers are possible (a) if numbers can be repeated and (b) if numbers cannot be repeated?

a. There are 10 choices for each digit. Use the fundamental counting principle to find the total amount of phone numbers.

Phone numbers 5 10 p 10 p 10 p 10 5 10,000

b. If you cannot repeat digits, there are still 10 choices for the first number, but then only 9 remaining choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit. Use the fundamental counting principle.

Phone numbers 5 10 p 9 p 8 p 7 5 5040

Example 2 Use the counting principle with repetition

1. If the pizza crust was not a choice in Example 1, how many different pizzas could be made?

160


Playoffs Eight teams are competing in a baseball playoff.

a. In how many different ways can the baseball teams finish the competition?

b. In how many different ways can 3 of the baseball teams finish first, second, and third?

Solutiona. There are 8! different ways that the teams can finish.

8! 5 8 p 7 p 6 p 5 p 4 p 3 p 2 p 1 5 40,320

b. Any of the 8 teams can finish first, then any of the 7 remaining teams can finish second, and then any of the remaining 6 teams can finish third.

8 p 7 p 6 5 336

Example 3 Find the number of permutations


Homework

Your NotesPERMUTATIONS OF n OBJECTS TAKEN r AT A TIME

The number of permutations of r objects taken from a group of n distinct objects is denoted by nPr .

nPr 5 n! } (n 2 r)!

Homework You have 6 homework assignments to complete over the weekend. However, you only have time to complete 4 of them on Saturday. In how many orders can you complete 4 of the assignments?

SolutionFind the number of permutations of 6 objects taken 4 at a time.

6P4 5 !

( )! 5 !

! 5 5

You can complete the 4 assignments in different orders.

Example 4 Find permutations of n objects taken r at a time

2. How many different 7 digit telephone numbers are possible if all of the digits can be repeated?

3. In Example 3, how many different ways can the teams finish if there are 6 teams competing in the playoffs?

4. You were left a list of 9 chores to complete. In how many orders can you complete 5 of the chores?



Homework

Your NotesPERMUTATIONS OF n OBJECTS TAKEN r AT A TIME

The number of permutations of r objects taken from a group of n distinct objects is denoted by nPr .

nPr 5 n! } (n 2 r)!

Homework You have 6 homework assignments to complete over the weekend. However, you only have time to complete 4 of them on Saturday. In how many orders can you complete 4 of the assignments?

SolutionFind the number of permutations of 6 objects taken 4 at a time.

6P4 5 6 !

6 2 4( )! 5

6 !

2 ! 5

2

720 5 360

You can complete the 4 assignments in 360 different orders.

Example 4 Find permutations of n objects taken r at a time

2. How many different 7 digit telephone numbers are possible if all of the digits can be repeated?

10,000,000

3. In Example 3, how many different ways can the teams finish if there are 6 teams competing in the playoffs?

720

4. You were left a list of 9 chores to complete. In how many orders can you complete 5 of the chores?

15,120



Homework

Your NotesPERMUTATIONS WITH REPETITION

The number of distinguishable permutations of n objects where one object is repeated s1 times, another is repeated s2 times, and so on, is:

n! }} s1! p s2 p . . . p sk!

Find the number of distinguishable permutations of the letters in (a) EVEN and (b) CALIFORNIA.

Solution

a. EVEN has letters of which is repeated times. So, the number of distinguishable

permutations is !

! 5 5

b. CALIFORNIA has letters of which and are each repeated times. So, the number of distinguishable permutations is

5 5 .

Example 5 Find permutations with repetition

5. TOMORROW

6. YESTERDAY

Checkpoint Find the number of distinguishable permutations of the letters in the word.


Homework

Your NotesPERMUTATIONS WITH REPETITION

The number of distinguishable permutations of n objects where one object is repeated s1 times, another is repeated s2 times, and so on, is:

n! }} s1! p s2 p . . . p sk!

Find the number of distinguishable permutations of the letters in (a) EVEN and (b) CALIFORNIA.

Solution

a. EVEN has 4 letters of which E is repeated 2 times. So, the number of distinguishable

permutations is 4 !

2 ! 5

2

24 5 12

b. CALIFORNIA has 10 letters of which A and I are each repeated 2 times. So, the number of distinguishable permutations is

10! } 2! p 2! 5

3,628,800 } 2 p 2 5 907,200 .

Example 5 Find permutations with repetition

5. TOMORROW

3360

6. YESTERDAY

90,720

Checkpoint Find the number of distinguishable permutations of the letters in the word.



Use Combinations and the Binomial TheoremGoal p Use combinations and the binomial theorem.

Your Notes VOCABULARY

Combination

Pascal’s triangle

Binomial theorem

COMBINATIONS OF n OBJECTS TAKEN r AT A TIME

The number of combinations of r objects taken from a group of n distinct objects is denoted by nCr.

nCr 5 n! } (n 2 r)! p r!

Books You are picking 7 books from a stack of 32. If the order of the books you choose is not important, how many different 7 book groups are possible?

The number of ways to choose 7 books from 32 is:

32C7 5 !

! p !

5

5

Example 1 Find combinations

10.6


Use Combinations and the Binomial TheoremGoal p Use combinations and the binomial theorem.

Your Notes VOCABULARY

Combination A selection of r objects from a group of n objects where the order is not important

Pascal’s triangle An arrangement of the values of nCr in a triangular pattern in which each row corresponds to a value of n

Binomial theorem For any positive integer n, the binomial expansion of (a 1 b)n is

(a 1 b)n 5 nC0anb0 1 nC1an 2 1b1 1 nC2an 2 2b2

1 . . . 1 nCna0bn

COMBINATIONS OF n OBJECTS TAKEN r AT A TIME

The number of combinations of r objects taken from a group of n distinct objects is denoted by nCr.

nCr 5 n! } (n 2 r)! p r!

Books You are picking 7 books from a stack of 32. If the order of the books you choose is not important, how many different 7 book groups are possible?

The number of ways to choose 7 books from 32 is:

32C7 5 32 !

25 ! p !7

5 32 p 31 p 30 p 29 p 28 p 27 p 26 p 25! }}}} 25! p 7!

5 3,365,856

Example 1 Find combinations

10.6

Your Notes

Movie Rentals The local movie rental store is having a special on new releases. The new releases consist of 12 comedies, 8 action, 7 drama, 5 suspense, and 9 family movies.

a. You want exactly 2 comedies and 3 family movies. How many different movie combinations can you rent?

b. You can afford at most 2 movies. How many movie combinations can you rent?

Solutiona. You can choose 2 of the 12 comedies and 3 of the

9 family movies. So, the number of possible sets of movies is:

12C2 p 9C3 5 !

! ! p !

! !

5 5

b. You can rent 0, 1, or 2 movies. Because there are movies to choose from, the number of possible

sets of movies is:

C0 1 C1 1 C2 5 5

Example 2 Decide to multiply or add combinations

1. Find 7C4. 2. Find 6C3. 3. Find 12C11.

4. From Example 2, find the number of possible movie combinations if you can choose 2 action movies and 2 dramas.



Your Notes

Movie Rentals The local movie rental store is having a special on new releases. The new releases consist of 12 comedies, 8 action, 7 drama, 5 suspense, and 9 family movies.

a. You want exactly 2 comedies and 3 family movies. How many different movie combinations can you rent?

b. You can afford at most 2 movies. How many movie combinations can you rent?

Solutiona. You can choose 2 of the 12 comedies and 3 of the

9 family movies. So, the number of possible sets of movies is:

12C2 p 9C3 5 12 !

10 ! !2 p

9 !

6 ! !3

5 66 p 84 5 5544

b. You can rent 0, 1, or 2 movies. Because there are 41 movies to choose from, the number of possible sets of movies is:

41 C0 1 41 C1 1 41 C2 5 1 1 41 1 820 5 862

Example 2 Decide to multiply or add combinations

1. Find 7C4. 2. Find 6C3. 3. Find 12C11.

35 20 12

4. From Example 2, find the number of possible movie combinations if you can choose 2 action movies and 2 dramas.

588



Homework

Your Notes

Reading A popular magazine has 11 articles. You want to read at least 2 of the articles. How many different combinations of articles can you read?

SolutionFor each of the 11 articles, you can choose to read or not read the article, so there are total combinations. If you read at least articles, you do not read only a total of or articles. So, the number of ways you can read at least 2 articles is:

2 (11C 1 11C ) 5 5

Example 3 Solve a multi-step problem

5. Your school football team has 10 scheduled games for the season. You want to attend at least 4 games. How many different combinations of games can you attend?


PASCAL’S TRIANGLE

The first and last numbers in each row are . Every number other than is the sum of the closest two numbers in the row directly above it.

Pascal’s triangle: As combinations

n 5 0 (0th row) 0C

n 5 1 (1st row) 1C 1C

n 5 2 (2nd row) 2C 2C 2C

n 5 3 (3rd row) 3C 3C 3C 3C

As numbers


Homework

Your Notes

Reading A popular magazine has 11 articles. You want to read at least 2 of the articles. How many different combinations of articles can you read?

SolutionFor each of the 11 articles, you can choose to read or not read the article, so there are 211 total combinations. If you read at least 2 articles, you do not read only a total of 0 or 1 articles. So, the number of ways you can read at least 2 articles is: 211 2 (11C 0 1 11C 1 ) 5 2048 2 (1 1 11) 5 2036

Example 3 Solve a multi-step problem

5. Your school football team has 10 scheduled games for the season. You want to attend at least 4 games. How many different combinations of games can you attend?

792


PASCAL’S TRIANGLE

The first and last numbers in each row are 1 . Every number other than 1 is the sum of the closest two numbers in the row directly above it.

Pascal’s triangle: As combinations

n 5 0 (0th row) 0C 0

n 5 1 (1st row) 1C 0 1C 1

n 5 2 (2nd row) 2C 0 2C 1 2C 2

n 5 3 (3rd row) 3C 0 3C 1 3C 2 3C 3

As numbers

1

1 1

1 2 1

1 3 3 1


Homework

Your Notes


Class Representatives Out of 5 finalists, your class must choose 3 class representatives. Use Pascal’s triangle to find the number of combinations of 3 students that can be chosen as representatives.

Solution

Find 5C using the 5th row of Pascal’s triangle.

n 5 5(5th row)

5C 5C 5C 5C 5C 5C

The value of 5C is the value in the 5th row of Pascal’s triangle. Therefore, 5C 5 . There are combinations of class representatives.

Example 4 Use Pascal's triangle

6. In Example 4, use Pascal's triangle to find the number of combinations of 3 students that can be chosen from 8 finalists.


BINOMIAL THEOREM

• For any positive integer n, the binomial expansion of (a 1 b)n is:

(a 1 b)n 5 nC0anb0 1 nC1an 2 1b1 1 . . . 1 nCna0bn

Notice that each term in the expansion of (a 1 b)n has the form where r is an integer from 0 to n.

Use the binomial theorem to write the binomial expansion.

(x 1 4)3

5

5

5

Example 5 Expand a power of a binomial sum

Homework

Your Notes


Class Representatives Out of 5 finalists, your class must choose 3 class representatives. Use Pascal’s triangle to find the number of combinations of 3 students that can be chosen as representatives.

Solution

Find 5C 3 using the 5th row of Pascal’s triangle.

n 5 5(5th row)

1 5 10 10 5 1 5C 0 5C 1 5C 2 5C 3 5C 4 5C 5

The value of 5C 3 is the fourth value in the 5th row of Pascal’s triangle. Therefore, 5C 3 5 10 . There are 10 combinations of class representatives.

Example 4 Use Pascal's triangle

6. In Example 4, use Pascal's triangle to find the number of combinations of 3 students that can be chosen from 8 finalists.

56


BINOMIAL THEOREM

• For any positive integer n, the binomial expansion of (a 1 b)n is:

(a 1 b)n 5 nC0anb0 1 nC1an 2 1b1 1 . . . 1 nCna0bn

Notice that each term in the expansion of (a 1 b)n has the form nCran 2 rbr where r is an integer from 0 to n.

Use the binomial theorem to write the binomial expansion.

(x 1 4)3

5 3C0x3(4)0 1 3C1x2(4)1 1 3C2x1(4)2 1 3C3x0(4)3

5 (1)(x3)(1) 1 (3)(x2)(4) 1 (3)(x)(16) 1 (1)(1)(64)

5 x3 1 12x2 1 48x 1 64

Example 5 Expand a power of a binomial sum

Words to ReviewGive an example of the vocabulary word.

Sequence

Series

Arithmetic sequences

Arithmetic series

Common ratio

Terms

Summation (Sigma) notation

Common difference

Geometric sequences

Geometric series

284 Words to Review • Algebra 2 Notetaking Guide Copyright © Holt McDougal. All rights reserved.

Words to ReviewGive an example of the vocabulary word.

Sequence

an 5 2n 2 3

Series

21 1 1 1 3 1 5 1 7 1 . . .

Arithmetic sequences

26, 23, 20, 17, 14, . . . or an 5 29 2 3n

Arithmetic series

i 5 1

∑ 35

1 1 6i

Common ratio

For an 5 2(5)n 2 1, r 5 5

Terms

For an 5 2n 2 321, 1, 3, 5, 7, . . .

Summation (Sigma) notation

i 5 1

∑ 8

2i 2 3

Common difference

For an 5 29 2 3n, d 5 23

Geometric sequences

2, 10, 50, 250, 1250, . . .or an 5 2(5)n 2 1

Geometric series

k 5 1

∑ 12

6(2)k 2 1

284 Words to Review • Algebra 2 Notetaking Guide Copyright © Holt McDougal. All rights reserved.

Partial sum

Factorial

Pascal's triangle

Permutation

Combination

Copyright © Holt McDougal. All rights reserved. Words to Review • Algebra 2 Notetaking Guide 285

Review your notes and Chapter 10 by using the Chapter Review on pages 730–733 of your textbook.

Partial sum

For 3 } 4 , 9 }

16 , 27 }

64 , 81 }

256 , . . .;

S2 5 21 } 16

Factorial

There are 4! 5 24 ways to arrange the letters A, E, G, and T.

Pascal's triangle

The third row: 1 3 3 1So, 3C2 5 3.

Permutation

GATE is a permutation of the letters A, E, G, and T.

Combination

9C2 5 36

Copyright © Holt McDougal. All rights reserved. Words to Review • Algebra 2 Notetaking Guide 285

Review your notes and Chapter 10 by using the Chapter Review on pages 730–733 of your textbook.

Documents

10.1 Define and Use Sequences and Series€¦ · Define and Use Sequences and Series Goal p Recognize and write rules for number patterns. VOCABULARY Sequence A function whose domain