chapter 11: test your proficiency

Chapter 11: Test Your Proficiency

Directions: •Select a section to work on.•Work out each problem on a piece of paper. •Click to check your answer.•For detailed steps click on the provided link.•Move on to the next problem or return to the menu.

Sections

11-1: The Arithmetic Sequence

11-2: The Arithmetic Series

11-3: The Geometric Sequence

11-4: The Geometric Series

11-5: The Infinite Geometric Series

11-7: The Binomial Theorem

11-8: Mathematical Induction

11-6: Recursion and Iteration

I love sequences and

series!

Applications/Word Problems

11-1: The Arithmetic SequenceGo on to the next problem

Check Answer

1. Find the 10th term of the arithmetic sequence

9,13,17, Use the correct formula.

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Click here if you would you like to see a detailed explanation.

10 45a



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Go on to the next problem


Step 1: Recall the formula to find a term of an arithmetic sequence. 1 1na a n d Step 2: Identify the known values and plug them into the formula.

10

1

1

0

4

1

9 9

10

3 4

?

10

9

1

a

a

a

n

d

Step 3: Simplify and state the answer.

10

10

10

1

9 9 4 9 36

4

109 4

5

a

a

a



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Check Answer

2. Find the three arithmetic means between

18 and 46. Use the correct formula.

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The arithmetic means are 25,32,39.




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Step 1: Set up the problem and recall the formula to find a term of an arithmetic sequence.

118, ____, ____, ____,46 formula : 1 na a n dStep 2: Identify the known values and plug them into the formula.

1 18 1

46

46 8 5 1

?

5

na

a d

d

n

Step 3: Solve for d. Then find the missing arithmetic means by adding d to each preceding term.

46 18 5 1

46 18 4

28 4

7

d

d

d

d

2

3

4

18 7 25

25 7 32

32 7 39

a

a

a

The arithmetic means are 25,32,39.



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Check Answer

3. 230 is the ____ term of the arithmetic sequence

10, 2, 14, ? Use the correct formula.

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It is the 21st term.




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Step 1: Recall the formula to find a term of an arithmetic sequence. 1 1na a n d

Step 2: Identify the known values and plug them into the formula.

1

12

2 10 12

230 10

10

30 12

?

n n

n

d

a

a

Step 3: Solve for the missing value and state the answer.

230 10 1 12

240 12 1

20 1

21

n

n

n

n




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Check Answer


10,4,18, ? Use the correct formula.

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Step 1: Recall the formula to find a term of an arithmetic sequence. 1 1na a n d

Step 2: Identify the known values and plug them into the formula.

1

1410

4 10

1

270 270

4

1

1

0

?n n

n

a

a

d

Step 3: Solve for the missing value and state the answer.

270 10 1 14

280 14 1

20 1

21

n

n

n

n




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Check Answer



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27 166a



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Step 1: Recall the formula to find a term of an arithmetic sequence. 1 1na a n d Step 2: Identify the known values and plug them into the formula.

1 27

27

16 1

27

27 1

?

7

9 1

6

6 7

a

a

a

d

n

Step 3: Simplify and state the answer.

27

27

27

1

16 26

16

7 16 182

1

727

66

a

a

a



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11-2: The Arithmetic SeriesGo on to the next problem

1. Find the sum of the first 12 terms of the arithmetic series

21 29 37 .

Check Answer

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21 29 37 .

The sum is -780.


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21 29 37 .

Step 1: Recall the formula to find the sum of an arithmetic series. 12n nn

S a a

Step 2: Identify the known values and unknown values.

1 1221 12 ? 8 ?na n a d S

Step 3: Find the value of an.

1 1 21 12 1

109

8n n

n

a a n d a

a

Step 4: Find Sn.

12 12

12

1221 109 6 130

2780

S S

S

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n

1

2. Find S for the arithmetic series in which

8, 161,and 52.na a n

Check Answer

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The sum is -4394.


n

1


8, 161,and 52.na a n

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S a a


1 528 52 161 ?na n a S

Step 3: Plug in the values to the formula and find Sn.

52 52

52

528 161 26 169

24394

S S

S

n

1


8, 161,and 52.na a n

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3. Find the sum of the arithmetic series -16 + -11 + -6 + + 54.

Check Answer

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The sum is 285.



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S a a


1 16 ? 54 11 ( 16) 5 ?n na n a d S

Step 3: Find the value of n.

1Use 1 54 16 1 5

70 5 1

14 1

15

na a n d n

n

n

n

Step 4: Find Sn.

15 15

15

1516 54 7.5 38

2285

S S

SReturn

to Menu



18

3

4. Find the sum of the arithmetic series 3 4 . n

n

Check Answer

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The sum is 568.


18

3


n

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S a a

Step 2: Find a1.

1

1

For the value of (the first n value).

So, a

n is 3

33 4 13

a

18

3


n

Step 3: Find an.

Step 4: Find the value of n.

n

For the value of (the last n value).

So, a 3

n is 18

18 4 58

na

For the value of

18 3 1 16

(always subtract the last and first n values and ad )

n:

d 1

n

Click to see the rest of the explanation.

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Step 5: Plug in the values and find Sn.

16 16

16

1613 58 8 71

2568

S S

S

18

3


n

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11-3: The Geometric SequenceGo on to the next problem

1

1. Find the ninth term of the geometric sequence for which

2a 20 and . (Use the appropriate formula.)

3r

Check Answer

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1



3r

95120

6561a

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Step 1: Recall the formula to find a term of a geometric sequence. 11

nna a r

Step 2: Plug in the given values.

Step 3: Simplify.

Step 4: State the answer.

1



3r

9 1

92

Note: n = 9 (for ninth term) 203

a

8

9 8

20 2 20 256 5120

1 1 6561 65613a

95120

6561a

Note: Do not use rounded decimals unless the directions tell you to.

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1

2. Find the eleventh term of the geometric sequence for which

a 13 and 3. (Use the appropriate formula.)r

Check Answer

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11 767,637a

1



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Step 1: Recall the formula to find a term of a geometric sequence. 11

nna a r

Step 2: Plug in the given values.

Step 3: Simplify.


11 111Note: n = 11 (for eleventh term) 13 3a

1011 13 3 767,637a

Note: always enclose negative numbers in parentheses when raising to a power

1



11 767,637a

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Check Answer

27

3. Find the next three terms of the geometric sequence 6, 9, , .2

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81 243 729, ,

4 8 16

27


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Strategy: Use the definition of a geometric sequence: Each term after the first is found by multiplying the previous term by the common ration.

Step 1: Find the value of r (the common ratio).

Step 2: Find the next three terms in the sequence. Step 3: State the answer.

2

1

9 3Recall:

6 2

ar r

a

4 3

5 4

6 5

27 3 81

2 2 4

81 3 243

4 2 8

243 3 729

8 2 16

a a r

a a r

a a r

27


81 243 729, ,

4 8 16

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11-4: The Geometric SeriesGo on to the next problem

Check Answer

7 7 14

1. Find the sum of the geometric series to 8 terms.6 3 3

Use the appropriate formula.

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81 595

297 or 2 2

S


7 7 14



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Strategy: Use the formula to find the sum of a geometric series with a finite number of terms, n.

Step 1: Find the value of r (the common ratio), the value of a1, and the value of n.

Step 2: Plug in r, a1 and n to the formula. Step 3: State the answer.

21

1

77 6 73 2 8 (the number of terms being added)

7 3 7 66

ar r a n

a

8

8

7 72

6 61 2

S


7 7 14



81 595

297 or 2 2

S

1 1Recall:

1

n

na a r

Sr

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Check Answer

16

n=1

12. Find the sum of the geometric series 5 . Use the appropriate formula.

2

n

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627 315

9 or 32 32

S


16

n=1


2

n

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Step 1: Identify the value of r (the common ratio), the value of a1, and the value of n from the sigma notation.

Step 2: Plug in r, a1 and n to the formula.


6

6

15 5

21

12

S


1 1Recall:

1

n

na a r

Sr

16

n=1


2

n

627 315

9 or 32 32

S

1

6 11

52

n

1a r

6 1 1

6

n

n

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Check Answer

13. Find the sum of the geometric series for which 19, 4, and 10.


a r n

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10 3,984,585S




a r n

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Step 1: Plug in r, a1 and n to the formula. Step 2: State the answer.

10

1019 19 4

1 4S


1 1Recall:

1

n

na a r

Sr

10 3,984,585S

Note: always enclose negative numbers in parentheses when raising to a power



a r n

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11-5: The Infinite Geometric SeriesGo on to the next problem

Check Answer

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


8 20 50

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51

2r




8 20 50

The sum does not exist.

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Recall that an infinite geometric series only has a sum under certain conditions. The value of the common ratio, r, must be within the interval -1 < r < 1. Otherwise, the sum does not exist.

Step 1: Find the value of r. Step 2: Compare the value of r to the interval -1 < r < 1.

2

1

20 5

8 2

ar

a

51

2r




8 20 50

The value of r is outside of the interval.

Step 3: Do not use the sum formula. Make the concluding statement.

51

2r The sum does not exist.

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Check Answer



2512 10

3

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72S




2512 10

3

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Step 1: Find the value of r. Step 2: Compare the value of r to the interval -1 < r < 1.

2

1

10 5

12 6

ar

a

51 1

6


The value of r is inside of the interval.

Step 3: Use the sum formula for an infinite geometric series.



2512 10

3

1 12 1212 6 72

5 11 16 6

aS

r

Step 4: Make the concluding statement.

72S

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Check Answer

1

1


2 Use the appropriate formula. 3

3

n

n

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9S


1

1



3

n

n

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Step 1: Identify the value of r from the sigma notation.

Step 2: Compare the value of r to the interval -1 < r < 1.

2

1 13


The value of r is inside of the interval.

Step 3: Use the sum formula for an infinite geometric series. Note: here a1 is 3

12

3

33 3

13

39

11

aS

r


9S

1

1



3

n

n

1

1

23

3

n

nr

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11-6: Recursion and IterationGo on to the next problem

Check Answer

1

1

1. Find the first five terms of the sequence where 7

and 3 8.n n

a

a a

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7,13,31,85,247


1

1


and 3 8.n n

a

a a

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Step 1: Always begin with n = 1.

Step 2: Now plug in the value of .

Step 3: Repeat this process with n = 2, then n = 3, and then n = 4.


1

1


and 3 8.n n

a

a a

7,13,31,85,247

1 1

1

1 1

2

If , then 3 8 becomes 3 8

which yields

n = 1

3 8n na a a a

a a

1a

12 23 8 become 1s 3 7a 8 3a a

1

1

1

2

3

2 3 3

4 4

54 54

2

3 3

4

If , then 3 8 3 8 3 8

If , then 3 8 3

n = 2

n = 8 3 8

If , then 3 8

13

31

8

3

n =

31

85

3 2484 5 73 8

a a a a

a a a a

a a a

a

a

a

a

Step 4: State the first five terms in sequence.

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Check Answer

1 2

2 1

2. Find the first five terms of the sequence where 9, 15

and 2 .n n n

a a

a a a

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9,15, 3, 33, 27


1 2

2 1


and 2 .n n n

a a

a a a

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Step 1: Always begin with n = 1.

Step 2: Now plug in the values of . Be careful to place the values correctly.

Step 3: Repeat this process with n = 2 and n = 3.


1

2 1 11 2

3

1

2

1If , then 2 becomes 2

which yield

n

s 2

= 1 n n na a

a a

a a a

a

a

12 an d a a

23 1 32 become 3s 15a 2 9a a a

2

3

4 42 2 2

3 3

2 1

2 451

3

3 5

If , then 2 2 a 2

I

1

f ,

n 3

then

3

27

3

5

3

= 2

n 2 3= 2 23 a 3

a

a

aa a a

a

a

a a a a

Step 4: State the first five terms in sequence.

1 2

2 1


and 2 .n n n

a a

a a a

9,15, 3, 33, 27 Return to

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Check Answer

03. Find the first three iterates of 9 2 for an initial value of x 5.f x x

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43, 385, 3463



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Step 1: Find the first iterate, x1 by finding f (x0).

Step 2: Find the second iterate, x2, by finding f (x1).


1 0 5 2 39 5 4xfx f

Step 4: State the first three iterates in sequence.


43, 385, 3463

12 943 4 383 52f fxx

Step 3: Find the third iterate, x3, by finding f (x2).

23 385 39 2 385 463xfx f

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Check Answer


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46, 6346, 120,815,146



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Step 1: Find the first iterate, x1 by finding f (x0). Remember to follow the order of operations.

Step 2: Find the second iterate, x2, by finding f (x1).


201 4 4 63 42xx f f

Step 4: State the first three iterates in sequence.

122

3 2 64 46 466 3xfx f

Step 3: Find the third iterate, x3, by finding f (x2).

232

6346 63 120,3 82 1 , 66 144 5f x fx


46, 6346, 120,815,146

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11-7: The Binomial TheoremGo on to the next problem

Check Answer

51. Expand 5 2 using the Binomial Theorem.x

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5 5 4 3 25 2 3125 6250 5000 2000 400 32x x x x x x

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Note: These problems are much easier if you learn the pattern from term to term rather than try to memorize the Binomial Theorem as a formula.

Step 1: Write out the expansion.

Note the patterns: 1) the powers on the first term (5x) decrease (5, 4, 3, 2, 1, 0)2) The powers on the second term (2) increase (0,1, 2, 3, 4, 5)3) The factorial increases (0!, 1!, 2!, 3!, 4!, 5!)4) The previous power of (5x) moves to the front as a multiplier in the next term.

5 4 3 25

1 2 3

4 51

0! 1

5 5 4 55 5 2 5 2 5 25 2

5 2

4 3

5 4 3 2 5 4 3

! 2!

2

3!

4

1 2

! 5!

x x x xx

x

Continue to the next slide to see the rest of the explanation.



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Step 2: Simplify the terms in the expansion. Remember to follow the order of operations.

5 4 3 25

3125 625 2 125 4 25 8

1 1 2

5 5 4 5 4 3

5 4 3 2 5 4 3 2

5 2

5 16 11 3 2 1

4 3 2 1 5 3

32

4 2 1

x x x xx

x



5 4 3 25

3125 625 2 125 4 25 8

1 1 2

5 5 4 5 4 3

5 4 3 2 5 4 3 2

5 2

5 16 1

1 3 2 1

4 3 2 1 5 3

32

4 2 1

x x x xx

x

Step 3: Reduce as much as possible.2 2

Return to the previous slide


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5 5 4 3 25 2 3125 6250 5000 2000 400 32x x x x x x


Step 4: Simplify each numerator and write the final expansion.

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Check Answer

42. Expand 4 3 using the Binomial Theorem.a b

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4 4 3 2 2 3 44 3 256 768 864 432 81a b a a b a b ab b


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Note: These problems are much easier if you learn the pattern from term to term rather than try to memorize the Binomial Theorem as a formula.

Step 1: Write out the expansion. Be sure to include the negative sign.

Note the patterns: 1) the powers on the first term (4a) decrease (4, 3, 2, 1, 0)2) The powers on the second term (-3b) increase (0,1, 2, 3, 4)3) The factorial increases (0!, 1!, 2!, 3!, 4!)4) The previous power of (4a) moves to the front as a multiplier in the next term.

4 3 1 2

3

42

1 4

4 4 3 4 34 3

4 3

4

0! 1! 2!

3

4 3

4 3 2 4 2

! 4!

23 1

a a b a ba b

a b




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Step 2: Simplify the terms in the expansion. Remember to follow the order of operations.


Step 3: Reduce as much as possible.2

2




4 3 2 24

3

1 1 2 1

3 2

256 64 3 164 4 3

4 3 2 4 3

94 3

4 27 16

1 4 1

2

2

1

3

a a b a ba b

a b

4 3 2 24

3

1 1 2 1

3 2

256 64 3 164 4 3

4 3 2 4 3

94 3

4 27 16

1 4 1

2

2

1

3

a a b a ba b

a b

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11-7: The Binomial TheoremReturn to the previous slide

Step 4: Simplify each numerator and write the final expansion.


4 4 3 2 2 3 44 3 256 768 864 432 81a b a a b a b ab b

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Check Answer

7

3. Use the Binomial Theorem (factorial formula) to find the fifth term in the

expansion of 3 5 .x

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3590,625x

7


expansion of 3 5 .x

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Step 1: Recall the factorial formula.

!Given

! !n k knn

na

k kb ba



7


expansion of 3 5 .x

Step 2: Identify the values for n, k, a, and b.

is the power on the binomial:

is one less than the location of the term: 5 1

a 3

k

n 7

b

is the first term in the binomial:

is the second term in the binom 5

4

ial:

x

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Step 3: Plug the values into the formula.

47 4!

! !73

7

4 45x


7


expansion of 3 5 .x

Step 4: Simplify.

3 4 37!3 5 35 27 625

3! 4!x x


3590,625x


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Check Answer

8

4. Use the Binomial Theorem (factorial formula) to find the fourth term in the

expansion of 2 9 .x y

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5 31,306,368x y

8



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Step 1: Recall the factorial formula.

!Given

! !n k knn

na

k kb ba



Step 2: Identify the values for n, k, a, and b.

k 3

is the power on the binomial:

is one less than the location of the term: 4 1

is the first term in the binomial:

is the second term in the binomial:

a 2

b 9

n

8

x

y

8



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Step 3: Plug the values into the formula.

8 3 3

23

8

!3

!

!89x y


Step 4: Simplify.

5 3 5 38!2 9 56 32 729

5! 3!x y x y



8



5 31,306,368x yReturn to

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11-8: Mathematical InductionGo on to the next problem

Check Answer

1. Use the method of Mathematical Induction to prove that the statement

11 17 23 6 5 3 8 is true for all positive integers.n n n

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1

1

2

2

1

1. Identify , ,

2. Show the statement is true for n = S 3 8 1 11 11

3. Assume that 11 17 23 6 5 3 8

4. Proof: 11 17 23 6 5 6 1 5 3 8 6 1 5

3 8 6 6 5

3 1

1 1

4

1:n na a and S

a

k k k

k k k k k

k k k

k k

1

n

11

1 3 11

1 3 1 8

Thus S is valid.k

k k

k k

S

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Step 1: Identify a1, an, and Sn. (This helps you begin organizing your proof.)


Step 2: Let n = 1 and show that the sum formula, Sn, will be equal to the first term, a1. (This shows the statement is true for the first value of n.)



17 23 is true for all positive i6 5 nt1 e3 g s1 8 er .n nn

1 is the first term:

is the nth term formula:

is the sum formula which the statement is equal 3

6 5

to:

1

8

1

n

n

a

a n

S n n

1 1If , then n = 1 S 3 8 1 11 1 1 11 a

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Step 3: Rewrite the problem with k instead of n. (This change of variables indicates the proof is moving to the next stage. Don’t forget to begin with the word “Assume.”)

Step 4a: Find the ak+1 term. To do this change the k to (k + 1) in the an term. (This step is not part of the proof and should be done on the side.)





Assume that 11 17 23 6 5 3 8k k k

1 16 5 6 5k k ka ak


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Step 4: Add ak+1 to both sides of the statement in step 3. (Don’t forget to begin with the word “Prove or Proof.”)

Step 5: Find Sk+1. To do this change the k to (k + 1) in the Sn formula. (This step is not part of the proof and should be done on the side.) Note: This step gives you the end goal during the rest of the proof.





6 1 5Proof: 11 17 23 6 5 3 18 6 5k kk k k

13 8 3 8 or 1 3 11 1 1k kS S kk kk k k


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Step 6: Use algebra to change the right side of the statement into Sk+1. Be sure to clearly show all of the steps.





2

2

1

Proof: 11 17 23 6 5 3 8

3 8 6 6 5

3 14 11

1 3 11

6 1 5

1 3

1

8

6 5

1

k

k k k

k k k

k k

k k

k

S

k

k

k


Thus Sn is valid.

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Applications/Word ProblemsGo on to the next problem

Check Answer

1. A patio is in the shape of a triangle. The longest row has 36 bricks.

Each row after that has two less bricks. If there are 15 rows, how many

bricks are in the last row?

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There are 8 bricks in the last row.

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Step 1: Determine whether the sequence is arithmetic or geometric and write out the first few terms.

2

1

The sequence is , because it involves subtraction.

The number of bricks in the (the longest row): 36

The number of bricks in the (2 less bricks)se : cond

arit

36

first r

hme

2 34

Th

t

e

o

ic

w

n

ro

r

w

umbe

a

a

3of bricks in the (2 less brickthir s): 3d row 4 2 32a


Step 2: Determine whether you are finding a term or a sum.

Based on the first few terms in Step 1, the number of bricks in the last

row would be the last term, not a sum.

Therefore, you need to find not .n na S





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Step 3: Find an.

15

n 1

15

2 2 l

Here , because there are of bricks, and

, because each row has thess bricks an th

n = 15 1

e previous row.

Recall: a 1

So, a 36 1

36 28 8

5 r

5 2

ows

1

a n d

a

d

Step 4: Write the concluding sentence.

There are 8 bricks in the last row.






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Check Answer


Each row after that has two less bricks. If there are 15 rows of bricks,

how many bricks were used to make the patio?

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To make the patio 330 bricks were used.




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2

1




arit

36

first r

hme

2 34

Th

t

e

o

ic

w

n

ro

r

w

umbe

a

a




Since each term gives only the number of bricks in that row,

the number of bricks used in the entire patio would be the sum, not a term.

Therefore, you need to find not .n nS a





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Step 3: Find an.

1

n

Recall: 2

Here , because there are of bricks, but the value of

a (the number of bricks in the last row) isn't known and must be found first.

Now, , becaus

n

e 2 2 leea

= 15 15

ch

ro

row has

w

s

s

n nn

S a a

d

n

15

n 1

15

than the previous row.

So, a is found as follows:

Recall: a 1

Plug in the values,

s bricks

1a 36 15

36 8 8

2

2

a n d

a

Applications/Word ProblemsReturn to the previous slide





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Step 4: Find Sn.

1

15

Now, 2

15 15Plug in the known values: 36 8 44 15 22 330

2 2

n nn

S a a

S






To make the patio, 330 bricks were used.

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Check Answer


Each row after that has two less bricks. How many bricks are in the tenth row?

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There are 18 bricks in the tenth row.



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2

1




arit

36

first r

hme

2 34

Th

t

e

o

ic

w

n

ro

r

w

umbe

a

a




10

Since each term gives the number of bricks in that row,

the number of bricks in the tenth row would be the tenth term, a .




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Step 3: Find an.

10

10

n 1

10

Here , for the , and , because each row has

than the previous row

2

.


Recall: a 1

Plug in the values, a 36 1

36 18 18

n = 10 tenth row

1

2

0

less bricks

2

a n d

a

d


There are 18 bricks in the tenth row.




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Check Answer

4. A machine is driving a post into the ground. On the first strike the post goes

12 inches into the ground. Each strike after the first the post goes only 50%

as far into the ground. How far in the ground will the post be after 6 strikes?

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5The post will be 23 inches in the ground after 6 strikes.

8

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2

1

3

The sequence is , because it involves multiplication.

The depth with the : 12 inches

The depth with the : 12 0.5 6 inches

The dep

second

th wit

geome

h the

tri

t

first st

hird st

st

rik

r

e

r

:

i

ik

c

.5

ke

e

6 0

a

a

a 3 inches


Step 2: Determine whether the problem is asking you to find a term or a sum.





The wording “after 6 strikes” means we are looking for the total, or sum, of the first 6 depths, S6. (Note: “on the 6th strike” would have been a6.)

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Step 3: Find Sn.

1

6

6

6

1n

Here , for the , and .

So, S is found as

n = 6 after 0.5 (

follows:

Recall: S1

5Plug in the values, S

s forix

12

strik 5

23

0%)

0.5

es

12

51 80.

na a r

r

r






5The post will be 23 inches in the ground after 6 strikes.

8Return

to Menu



Check Answer



as far into the ground. How far will the post go into the ground on the 4th strike?

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1The post will go 1 inches into the ground on the 4th strike.

2




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2

1

3




The dep

second

th wit

geome

h the

tri

t

first st

hird st

st

rik

r

e

r

:

i

ik

c

.5

ke

e

6 0

a

a

a 3 inches




The wording “on the 4th strike” means we are looking for fourth term, a4. (Note: “after 4 strikes” would have been S4.)




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Step 3: Find an.

1n 1

4

414

Here , for the , and .


Recall: a

1Plug

n = 4 on the fourth s

in the values, a

0.5 (for 50%)

0.5

tri

12 12

ke

na r

r






1The post will go 1 inches into the ground on the 4th strike.

2Return

to Menu



Check Answer


12 inches into the ground. Each strike after the first the post goes only 50% as far

into the ground. What is the total distance the post can be driven into the ground?

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The total distance the post can be driven into the ground is 24 inches.




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2

1

3




The dep

second

th wit

geome

h the

tri

t

first st

hird st

st

rik

r

e

r

:

i

ik

c

.5

ke

e

6 0

a

a

a 3 inches




The word “total” implies we need to find a sum. The absence of a number of strikes to stop at means this is an “infinite geometric series” problem.




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Step 3: Find S.

1

Here there is and .

So, is found as follows:

Recall: , where 1 11

12Since 1

no 0.5 (for 50%)

0. 1,

valu

241

S

S

50.5

e r

ar

S

n

r






The total distance the post can be driven into the ground is 24 inches.

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chapter 11: test your proficiency

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