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Welcome to Interactive Chalkboard
Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio Welcome to Interactive Chalkboard

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Lesson 11-1 Arithmetic Sequences Lesson 11-2 Arithmetic Series
Lesson 11-3 Geometric Sequences Lesson 11-4 Geometric Series Lesson 11-5 Infinite Geometric Series Lesson 11-6 Recursion and Special Sequences Lesson 11-7 The Binomial Theorem Lesson 11-8 Proof and Mathematical Induction Contents

Example 1 Find the Next Terms Example 2 Find a Particular Term
Example 3 Write an Equation for the nth Term Example 4 Find Arithmetic Means Lesson 1 Contents

Find the next four terms of the arithmetic sequence –8, –6, –4, ….
Find the common difference d by subtracting 2 consecutive terms. Now add 2 to the third term of the sequence and then continue adding 2 until the next four terms are found. +2 +2 +2 +2 –4 – Answer: The next four terms are –2, 0, 2 and 4. Example 1-1a

Find the next four terms of the arithmetic sequence 5, 3, 1, ….
Answer: The next four terms are –1, –3, –5 and –7. Example 1-1b

Construction The table below shows typical costs for a construction company to rent a crane for one, two, three, or four months. Assuming that the arithmetic sequence continues, how much would it cost to rent the crane for 24 months? Months Cost($) 1 75,000 2 90,000 3 105,000 4 120,000 Example 1-2a

Months Cost($) 1 75,000 2 90,000 3 105,000 4 120,000 Explore Since the difference between any two successive costs is $15,000, the costs form an arithmetic sequence with common difference 15,000. Plan You can use the formula for the nth term of an arithmetic sequence with and to find the cost for 24 months. Example 1-2a

Formula for the nth term
Solve Formula for the nth term Simplify. Answer: It would cost $420,000 to rent for 24 months. Example 1-2a

Examine. You can find the term of the sequence by adding 15,000
Examine You can find the term of the sequence by adding 15,000. From Example 2 on page 579 of your textbook, you know the cost to rent the crane for 12 months is $120,000. So, a12 through a24 are 240,000, 255,000, 270,000, 285,000, 300,000, 315,000, 330,000, 345,000, 360,000, 375,000, 390,000, 405,000, and 420,000. Therefore, $420,000 is correct. Example 1-2a

Answer: It would cost $180,000 to rent for 8 months.
Construction The table below shows typical costs for a construction company to rent a crane for one, two, three, or four months. Assuming that the arithmetic sequence continues, how much would it cost to rent the crane for 8 months? Months Cost($) 1 75,000 2 90,000 3 105,000 4 120,000 Answer: It would cost $180,000 to rent for 8 months. Example 1-2b

In this sequence, and Use the nth formula to write an equation.
Write an equation for the nth term of the arithmetic sequence –8, –6, –4, …. In this sequence, and Use the nth formula to write an equation. Formula for the nth term Distributive Property Simplify. Answer: An equation is . Example 1-3a

Write an equation for the nth term of the arithmetic sequence 5, 3, 1, ….
Answer: An equation is . Example 1-3b

Find the three arithmetic means between 21 and 45.
You can use the nth term formula to find the common difference. In the sequence 21, ___, ___, ___, 45, ..., and Formula for the nth term Subtract 21 from each side. Divide each side by 4. Example 1-4a

Now use the value of d to find the three arithmetic means.
+6 +6 +6 Answer: The arithmetic means are 27, 33, and 39. Check Example 1-4a

Find the three arithmetic means between 13 and 25.
Answer: The arithmetic means are 16, 19, and 22. Example 1-4b

End of Lesson 1

Example 1 Find the Sum of an Arithmetic Series
Example 2 Find the First Term Example 3 Find the First Three Terms Example 4 Evaluate a Sum in Sigma Notation Lesson 2 Contents

Find the sum of the first 20 even numbers, beginning with 2.
The series is Since you can see that and you can use either sum formula for this series. Method 1 Sum formula Simplify. Multiply. Example 2-1a

Answer: The sum of the first 20 even numbers is 420.
Method 2 Sum formula Simplify. Multiply. Answer: The sum of the first 20 even numbers is 420. Example 2-1a

Find the sum of the first 15 numbers, beginning with 1.
Answer: The sum of the first 15 numbers is 120. Example 2-1b

Radio A radio station is giving away money every day in the month of September for a total of $124,000. They plan to increase the amount of money given away by $100 each day. How much should they give away on the first day of September, rounded to the nearest cent? You know the values of n, Sn, and d. Use the sum formula that contains d. Example 2-2a

Distributive Property
Sum formula Simplify. Distributive Property Subtract 43,500. Divide by 30. Answer: They should give away $ the first day. Example 2-2a

Answer: The first question is worth $25,000.
Games A television game show gives contestants a chance to win a total of $1,000,000 by answering 16 consecutive questions correctly. If the value of each question is increased by $5,000, how much is the first question worth? Answer: The first question is worth $25,000. Example 2-2b

Find the first four terms of an arithmetic series in which
Step 1 Since you know and use Example 2-3a

Step 2 Find d. Step 3 Use d to determine Example 2-3a

Answer: The first four terms are 14, 17, 20, 23.
Example 2-3a

Find the first three terms of an arithmetic series in which and
Answer: The first three terms are 11, 16, 21. Example 2-3b

Method 1 Find the terms by replacing k with 3, 4, …, 10. Then add.
Evaluate Method 1 Find the terms by replacing k with 3, 4, …, 10. Then add. Example 2-4a

Answer: The sum of the series is 112.
Method 2 Since the sum is an arithmetic series, use the formula There are 8 terms, Answer: The sum of the series is 112. Example 2-4a

Evaluate Answer: 108 Example 2-4b

End of Lesson 2

Example 1 Find the Next Term Example 2 Find a Particular Term
Example 3 Write an Equation for the nth Term Example 4 Find a Term Given the Fourth Term and the Ratio Example 5 Find Geometric Means Lesson 3 Contents

Multiple-Choice Test Item
Find the missing term in the geometric sequence 324, 108, 36, 12, ___. A 972 B 4 C 0 D –12 Read the Test Item Since the sequence has the common ratio of Example 3-1a

To find the missing term, multiply the last given term by
Solve the Test Item To find the missing term, multiply the last given term by Answer: B Example 3-1a

Multiple-Choice Test Item
Find the missing term in the geometric sequence 100, 50, 25, ___. A 200 B 0 C 12.5 D –12.5 Answer: C Example 3-1b

Find the sixth term of a geometric sequence for which and
Formula for the nth term Multiply. Answer: The sixth term is 96. Example 3-2a

Find the fifth term of a geometric sequence for which and
Answer: The fifth term is 96. Example 3-2b

Write an equation for the nth term of the geometric sequence 5, 10, 20, 40, …. Formula for the nth term Answer: An equation is Example 3-3a

Write an equation for the nth term of the geometric sequence 2, 6, 18, 54, ….
Answer: An equation is . Example 3-3b

Find the seventh term of a geometric sequence for which and
First find the value of Formula for the nth term Divide by 4. Example 3-4a

Now find a7. Formula for the nth term Use a calculator. Answer: The seventh term is 1536. Example 3-4a

Find the sixth term of a geometric sequence for which and
Answer: The sixth term is 243. Example 3-4b

Find three geometric means between 3.12 and 49.92.
Use the nth term formula to find the value of r. In the sequence 3.12, ___, ___, ___, 49.92, a1 is 3.12 and a5 is Formula for the nth term Divide by 3.12. Take the fourth root of each side. Example 3-5a

There are two possible common ratios, so there are two possible sets of geometric means. Use each value of r to find three geometric means. Answer: The geometric means are 6.24, 12.48, and , or –6.24, 12.48, and –24.96. Example 3-5a

Find three geometric means between 12 and 0.75.
Answer: The geometric means are 6, 3, and 1.5, or –6, 3, and –1.5. Example 3-5b

End of Lesson 3

Example 1 Find the Sum of the First n Terms
Example 2 Evaluate a Sum Written in Sigma Notation Example 3 Use the Alternate Formula for a Sum Example 4 Find the First Term of a Series Lesson 4 Contents

Answer: Going back 8 generations, a person would have 510 ancestors.
Genealogy How many direct ancestors would a person have after 8 generations? Counting two parents, four grandparents, eight great- grandparents, and so on gives you a geometric series with Sum formula Use a calculator. Answer: Going back 8 generations, a person would have 510 ancestors. Example 4-1a

Answer: Going back 7 generations, a person would have 254 ancestors.
Genealogy How many direct ancestors would a person have after 7 generations? Answer: Going back 7 generations, a person would have 254 ancestors. Example 4-1b

Evaluate Method 1 Find the terms by replacing n with 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Then add. Example 4-2a

Since the sum is a geometric series, you can use the formula
Method 2 Since the sum is a geometric series, you can use the formula Sum formula Example 4-2a

Answer: The sum of the series is 12,285.
Simplify. Answer: The sum of the series is 12,285. Example 4-2a

Find the sum of a geometric series for which
Since you do not know the value of n, use the formula Example 4-3a

Answer: The sum of the series is 6666.
Alternate sum formula Simplify. Answer: The sum of the series is 6666. Example 4-3a

Find the sum of a geometric series for which
Answer: The sum of the series is 74.4. Example 4-3b

Find a1 in a geometric series for which and
Sum formula Subtract. Divide each side by –255. Answer: The first term of the series is 3. Example 4-4a

Find a1 in a geometric series for which and
Answer: The first term of the series is 1. Example 4-4b

End of Lesson 4

Example 1 Sum of an Infinite Geometric Series
Example 2 Infinite Series in Sigma Notation Example 3 Write a Repeating Decimal as a Fraction Lesson 5 Contents

Find the sum of , if it exists.
First, find the value of r to determine if the sum exists. the sum does not exist. Answer: The sum does not exist. Example 5-1a

Find the sum of , if it exists.
the sum exists. Example 5-1a

Now use the formula for the sum of an infinite geometric series.
Sum formula Simplify. Answer: The sum of the series is 2. Example 5-1a

Find the sum of each infinite geometric series, if it exists. a.
b. Answer: no sum Answer: 2 Example 5-1b

In this infinite geometric series,
Evaluate In this infinite geometric series, Sum formula Simplify. Example 5-2a

Answer: Thus, Example 5-2a

Write the repeating decimal as a sum.
Write as a fraction. Method 1 Write the repeating decimal as a sum. Example 5-3a

In this series, Sum formula Example 5-3a

Subtract. Simplify. Example 5-3a

Label the given decimal.
Method 2 Label the given decimal. Repeating decimal Multiply each side by 100. Subtract the second equation from the third. Divide each side by 99. Answer: Thus, Example 5-3a

Write as a fraction. Answer: Example 5-3b

End of Lesson 5

Example 1 Use a Recursive Formula
Example 2 Find and Use a Recursive Formula Example 3 Iterate a Function Lesson 6 Contents

Find the first five terms of the sequence in which and
Recursive formula Example 6-1a

Answer: The first five terms of the sequence are 5, 17, 41, 89, 185.
Recursive formula Answer: The first five terms of the sequence are 5, 17, 41, 89, 185. Example 6-1a

Find the first five terms of the sequence in which and
Answer: The first five terms of the sequence are 2, 8, 26, 80, 242. Example 6-1b

Biology Dr. Elliott is growing cells in lab dishes
Biology Dr. Elliott is growing cells in lab dishes. She starts with 108 cells Monday morning and then removes 20 of these for her experiment. By Tuesday the remaining cells have multiplied by 1.5. She again removes 20. This pattern repeats each day in the week. Write a recursive formula for the number of cells Dr. Elliott finds each day before she removes any. Let cn represent the number of cells at the beginning of the nth day. She takes 20 out, leaving cn – 20. The number the next day will be 1.5 times as much. So, Answer: Example 6-2a

Find the number of cells she will find on Friday morning.
On the first morning, there were 108 cells, so Recursive formula Example 6-2a

Answer: On the fifth day, there will 303 cells.
Recursive formula Answer: On the fifth day, there will 303 cells. Example 6-2a

b. Find the number of cells she will find on Saturday morning. Answer:
Biology Dr. Scott is growing cells in lab dishes. She starts with 100 cells Monday morning and then removes 30 of these for her experiment. By Tuesday the remaining cells have doubled. She again removes 30. This pattern repeats each day in the week. a. Write a recursive formula for the number of cells Dr. Scott finds each day before she removes any. b. Find the number of cells she will find on Saturday morning. Answer: Answer: 1340 Example 6-2b

To find the first iterate x1, find the value of the function for
Find the first three iterates x1, x2, and x3 of the function for an initial value of To find the first iterate x1, find the value of the function for Iterate the function. Simplify. Example 6-3a

To find the second iterate x2, substitute x1 for x.
Iterate the function. Simplify. Substitute x2 for x to find the third iterate. Iterate the function. Simplify. Example 6-3a

Answer: Therefore, 5, 14, 41, 122 is an example of a sequence generated using iteration.
Example 6-3a

Find the first three iterates x1, x2, and x3 of the function
Find the first three iterates x1, x2, and x3 of the function for an initial value of Answer: 5, 11, 23 Example 6-3b

End of Lesson 6

Example 1 Use Pascal’s Triangle Example 2 Use the Binomial Theorem
Example 3 Factorials Example 4 Use a Factorial Form of the Binomial Theorem Example 5 Find a Particular Term Lesson 7 Contents

1 5 10 10 5 1 Expand Write row 5 of Pascal’s triangle.
Use the patterns of a binomial expansion and the coefficients to write the expansion of Answer: Example 7-1a

Expand Answer: Example 7-1b

Expand The expression will have nine terms. Use the sequence to find the coefficients for the first five terms. Use symmetry to find the remaining coefficients. Example 7-2a

Answer: Example 7-2a

Evaluate 1 Answer: Example 7-3a

Binomial Theorem, factorial form
Expand Binomial Theorem, factorial form Let Example 7-4a

Simplify. Example 7-4a

Answer: Example 7-4a

Find the fourth term in the expansion of
First, use the Binomial Theorem to write the expression in sigma notation. In the fourth term, Example 7-5a

Answer: Simplify. Example 7-5a

Find the fifth term in the expansion of
Answer: Example 7-5b

End of Lesson 7

Example 1 Summation Formula Example 2 Divisibility
Example 3 Counterexample Lesson 8 Contents

Assume for a positive integer k.
Prove that Step 1 When, the left side of the given equation is 2(1) –1 or 1. The right side is 12 or 1. Thus, the equation is true for Step 2 Assume for a positive integer k. Step 3 Show that the given equation is true for Example 8-1a

Add to each side. Add. Simplify. Factor. Example 8-1a

This proves that is true for all positive integers n.
The last expression is the right side of the equation to be proved, where n has been replaced by Thus, the equation is true for Answer: This proves that is true for all positive integers n. Example 8-1a

When, we have , which is true. Thus, the equation is true for
Prove that . Answer: Step 1 When, we have , which is true. Thus, the equation is true for Step 2 Assume that for a positive integer k. Step 3 Show that the given equation is true for Example 8-1b

The last expression is the right side of the equation to be proved, where n has been replaced by Thus, the equation is true for This proves that is true for all positive integers n. Example 8-1b

Prove that is divisible by 5 for all positive integers n.
Step 1 When, Since 5 is divisible by 5, the statement is true for Step 2 Assume that is divisible by 5 for some positive integer k. This means that there is a whole number r such that Example 8-2a

Show that the statement is true for
Step 3 Show that the statement is true for Inductive hypothesis Add 1 to each side. Multiply each side by 6. Simplify. Subtract 1 from each side. Factor. Example 8-2a

Thus, the statement is true for This proves that is divisible by 5.
Since r is a whole number, is a whole number. Therefore is divisible by 5. Answer: Thus, the statement is true for This proves that is divisible by 5. Example 8-2a

Prove that is divisible by 9 for all positive integers n.
Answer: Step 1 When, Since 9 is divisible by 9, the statement is true for Step 2 Assume that is divisible by 9 for some positive integer k. Example 8-2b

Show that this is true for .
Step 3 Show that this is true for . Example 8-2b

Since r is a whole number,. is a whole number. Therefore,
Since r is a whole number, is a whole number. Therefore, is divisible by 9. Thus, the statement is true for This proves that is divisible by 9. Example 8-2b

Check the first few positive integers. 1 12 + 1 + 5 or 7 yes 2
Find a counterexample for the formula that is always a prime number for any positive integer n. Check the first few positive integers. n Formula prime? 1 or 7 yes 2 or 11 3 or 17 4 or 25 no Answer: The value is a counterexample for the formula. Example 8-3a

The value is a counterexample for the formula.
Find a counterexample for the formula that is always a prime number for any positive integer n. Answer: The value is a counterexample for the formula. Example 8-3b

End of Lesson 8

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