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# Honors Geometry Sequences and Series!

How to determine the degree of a sequence and find the nth term of a sequence.

by

Tweet## Lynn Ibarra

on 19 April 2013#### Transcript of Honors Geometry Sequences and Series!

Sequences and Series!! Finding the nth term of a sequence! Arithmetic vs Geometric Sequences!

Series! Sigma Notation! Review!! To find the degree of a sequence, begin by finding the

difference between the numbers. 2, 5, 8, 11, 14

To get from 2 to 5 you add 3

To get from 5 to 8 you add 3

To get from 8 to 11 you add 3

To get from 11 to 14 you add 3 3, 10, 29, 66, 127, 218

To get from 3 to 10 you add 7

to get from 10 to 29 you add 19

To get from 29 to 66 you add 37

To get from 66 to 127 you add 61

To get from 127 to 218 you add 91 If the number you add each time is the same, you are done. Yay!

If the number you add each time is different, take the numbers you add each time and repeat the process.

If you complete this step one time, the sequence is linear. If you complete this step twice, it is quadratic, three times, it is cubic, four times it is quartic and so on. For the sequence 2, 5, 8, 11, 14 we are finished because we added three each time. Since we only had to perform this step once, this is a linear sequence. For the sequence 3, 10, 29, 66, 127, 218 we added 7 then 19, then 37, then 61, then 91. So now we will repeat the process with the numbers 7, 19, 37, 61, and 91.

To get from 7 to 19 you add 12

To get from 19 to 37 you add 18

To get from 37 to 61 you add 24

To get from 61 to add 91 you add 30.

We are still not adding the same thing each time so we will repeat the process again using the numbers 12, 18, 24, 30.

To get from 12 to 18 you add 6

To get from 18 to 24 you add 6

To get from 24 to 30 you add 6.

We are now adding the same amount each time so we stop. We had to repeat this process three times so this sequence is cubic. Finding the degree

of a Sequence! 2 5 8 11 14

+3 +3 +3 +3

This is LINEAR!! We had to find

the differences once to make

them all the same. 3 10 29 66 127 220

+7 +19 +37 +61 +91

+12 +18 +24 +30

+6 +6 +6

This is CUBIC!!! We had to find the differences three times before the differences were the same. Linear 1. Find the difference between the numbers.

2. Find out what you would have to add or subtract to the difference to get your beginning value.

3. Write your formula for the nth term:

(difference)*n + (the # you found in step 2)

Example: 6, 8, 10, 12, 14

1. The difference is 2.

2. To get to 6 (our starting value) we would have to add 4.

3. nth term = 2n+4

Example: 4, 1, -2, -5, -8

1. The difference is -3.

2. To get to 4 we would have to add 7.

3. nth term = -3n+7

Quadratic 1. List the factors of each term in the sequence.

2. Look for a linear pattern that uses one factor from each term.

3. Check and see if the other factors for each term also create a linear pattern.

4. Write the formula for the nth term for each of the two factors and multiply them together.

5. If you do not see a linear pattern in the factors, try doubling the original terms in the sequence and factor those numbers. When you find the pattern and write out the nth term, don't forget to divide by two.

6. If multiplying by two doesn't work, try multiplying by three (or another logical number; if you have fifths, multiply by 5).

Example: 8, 15, 24, 35, 48, 63, 80

Factors:

1, 8 1, 15 1, 24 1, 35 1, 48 1, 63 1, 80

2, 4 3, 5 2, 12 5, 7 2, 24 3, 21 2, 40

3, 8 3, 16 7, 9 4, 20

4, 6 4, 12 5, 16

6, 8 8, 10

If we use the factors (2, 4) (3, 5) (4, 6) (5, 7) (6, 8) (7, 9) and (8, 10) we can see the first number in each set of factors creates a nice linear sequence: 2, 3, 4, 5, 6, 7, 8 the formula for the nth term for this sequence is n + 1. The second set of numbers also creates a nice linear pattern: 4, 5, 6, 7, 8, 9, 10. The formula for the nth term for these numbers is n + 3. (See the instructions for finding the nth term of a linear sequence if you need a reminder.) So the formula for this quadratic sequence is (n + 1)(n + 3). Note we are just multiplying the formulas for the factors to get the original sequence, just as we multiply the factors themselves to get the actual numbers.

Example 2: -8, 11, 42, 85, 140, 207, 286

Factors:

1, -8 1, 11 1, 42 1, 85 1, 140 1, 207 1, 286

2, -4 2, 21 5, 17 2, 70 3, 69 2, 143

-1, 8 3, 14 4, 35 9, 23 11, 26

-2, 4 6, 7 5, 28

7, 20

If we use the set of factors (-1, 8) (1, 11) (3, 14) (5, 17) (7, 20) (9, 23) and (11, 26) the first set of numbers in these factors go up by two each time and the other numbers go up by 3 each time so they are both linear. The formula for the first term is (2n - 3) and the formula for the second term is (3n + 5), so the formula for the quadratic sequence is (2n - 3)(3n + 5).

Example 3: -4, -3, -6/5, 7/5, 24/5, 9, 14

You'll notice right away that we have fifths in this example. The logical thing to do would be to multiply by 5 so that we have whole numbers, then factor. So we will find the factors of: -20, -15, -6, 7, 24, 45, 70

1, -20 1, -15 1, -6 1, 7 1, 24 1, 45 1, 70

2, -10 3, -5 2, -3 2, 12 3, 15 2, 35

4, -5 -3, 5 -2, 3 3, 8 5, 9 5, 14

-4, 5 -1, 15 -1, 6 4, 6 7, 10

-2, 10

-1, 20

If we use the sets (-5, 4) (-3, 5) (-1, 6) (1, 7) (3, 8) (5, 9) (7, 10) the first set of numbers increases by two each time and the second set of numbers increases by 1. The formula for the first set of numbers is (2n - 7) and the formula for the second set is (n + 3). The formula for the quadratic sequence then is ((2n - 7)(n + 3)) /5. It is very important you don't forget to divide by 5 at the end because we multiplied by 5 before finding factors!

Cubic We know that a cubic will have the form n^3 + n^2 + n with possible coefficients before each of these and perhaps adding a constant at the end: an^3 + bn^2 + cn + d. We will use this knowledge and the fact that if you put in 1 for n you get the first term in the sequence, if you put 2 in for n you get the second term in the sequence, etc. Cubics are best explained with an example. The following is a cubic sequence: 10, 34, 88, 184, 334.

So we know that if we put in 1 for n we will get 10 for an answer:

(1) a + b + c + d = 10 (note I put in 1 for n)

We also know that if we put 2 in for n, we will get 34:

(2) 8a + 4b + 2c + d = 34.

We will continue to do this because we know if we have 4 unknowns, we need four equations:

(3) 27a + 9b + 3c + d = 88

(4) 64a + 16b +4c +d = 184

Now I will use elimination to solve these equations. Notice that I have numbered each equation so that you can see which equations I am subtracting and given them new numbers so that I can refer to that line again later:

(5) = (2) - (1): 7a + 3b + c = 24

(6) = (3) - (2): 19a + 5b + c = 54

(7) = (4) - (3): 37a + 7b +c = 96

(8) = (6) - (5): 12a + 2b = 30

(9) = (7) - (6): 18a + 2b = 42

(10) = (9) - (8): 6a = 12

Now we can solve for a: a = 2.

Now we will use substitution to solve for b, c and d:

Substituting 2 in for a in equation (8) we find b = 3.

Substituting 2 in for a and 3 in for b in equation (5) we find c = 1.

Substituting 2 in for a, 3 in for b and 1 in for c in equation (1) we find d = 4.

The formula for this cubic sequence is 2n^3 + 3n^2 + n + 4.

Series vs. Sequences A sequence is simply a list of numbers which follow a pattern. A series, is when you have a list of numbers following a pattern and you are adding the numbers all together! An arithmetic sequence is a sequence with the difference between two consecutive terms constant. We have called these linear sequences in the past, either name is acceptable. The difference between terms is called the common difference (d).

To find the nth term of an arithmetic sequence you take the first term and add (n-1) times the common difference. t1 + (n-1)d A geometric sequence is a sequence with the ratio between two consecutive terms constant. You multiply by the same number each time. This ratio (or number you multiply by) is called the common ratio (r).

To find the nth term of a geometric sequence, you take the first term and multiply it by the common ratio to the (n-1) power. t1*r^(n-1) 10

Σ 2i

i=1 The symbol Σ is the capital Greek letter sigma and mathematically it means to add all the numbers up. The number under the Σ tells you what term to start with. In this case we are to start with the first term. The number above the Σ tells you when to stop. In this example, we are to stop when we reach the tenth term. The numbers to the right of the Σ tell us the actual equation we are using. So we would substitue 1 in for i and get 2, then we substitute 2 in for i and get 4, then we substitute 3 in for i and get 6 and we keep doing this until we substitute 10 in for i and get 20. Then we add all these numbers together: 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110 To find the sum of an arithmetic series, you use the following formula: S=(n/2)(a1+an), where S is the sum, n is the number of terms, a1 is the first term and an is the last term.

For example: 2+4+6+8+10+12+14+16+18+20

S=(5)(2+20) There are ten terms so n=10 the first term (a1) is 2 and the last term (an) is 20. Therefore S=110.

To find the sum of a geometric series, you use this formula: S=(a1(1-r^n))/(1-r), where S is the sum, a1 is the first term, r is the common ratio and n is the number of terms.

For example: 2+6+18+54+162+486

S=(2(1-3^6))/(1-3) = (2(1-729))/(1-3) = (2(-728))/(-2) = 728 Visit my website for a power point of review questions and answers!

Full transcriptSeries! Sigma Notation! Review!! To find the degree of a sequence, begin by finding the

difference between the numbers. 2, 5, 8, 11, 14

To get from 2 to 5 you add 3

To get from 5 to 8 you add 3

To get from 8 to 11 you add 3

To get from 11 to 14 you add 3 3, 10, 29, 66, 127, 218

To get from 3 to 10 you add 7

to get from 10 to 29 you add 19

To get from 29 to 66 you add 37

To get from 66 to 127 you add 61

To get from 127 to 218 you add 91 If the number you add each time is the same, you are done. Yay!

If the number you add each time is different, take the numbers you add each time and repeat the process.

If you complete this step one time, the sequence is linear. If you complete this step twice, it is quadratic, three times, it is cubic, four times it is quartic and so on. For the sequence 2, 5, 8, 11, 14 we are finished because we added three each time. Since we only had to perform this step once, this is a linear sequence. For the sequence 3, 10, 29, 66, 127, 218 we added 7 then 19, then 37, then 61, then 91. So now we will repeat the process with the numbers 7, 19, 37, 61, and 91.

To get from 7 to 19 you add 12

To get from 19 to 37 you add 18

To get from 37 to 61 you add 24

To get from 61 to add 91 you add 30.

We are still not adding the same thing each time so we will repeat the process again using the numbers 12, 18, 24, 30.

To get from 12 to 18 you add 6

To get from 18 to 24 you add 6

To get from 24 to 30 you add 6.

We are now adding the same amount each time so we stop. We had to repeat this process three times so this sequence is cubic. Finding the degree

of a Sequence! 2 5 8 11 14

+3 +3 +3 +3

This is LINEAR!! We had to find

the differences once to make

them all the same. 3 10 29 66 127 220

+7 +19 +37 +61 +91

+12 +18 +24 +30

+6 +6 +6

This is CUBIC!!! We had to find the differences three times before the differences were the same. Linear 1. Find the difference between the numbers.

2. Find out what you would have to add or subtract to the difference to get your beginning value.

3. Write your formula for the nth term:

(difference)*n + (the # you found in step 2)

Example: 6, 8, 10, 12, 14

1. The difference is 2.

2. To get to 6 (our starting value) we would have to add 4.

3. nth term = 2n+4

Example: 4, 1, -2, -5, -8

1. The difference is -3.

2. To get to 4 we would have to add 7.

3. nth term = -3n+7

Quadratic 1. List the factors of each term in the sequence.

2. Look for a linear pattern that uses one factor from each term.

3. Check and see if the other factors for each term also create a linear pattern.

4. Write the formula for the nth term for each of the two factors and multiply them together.

5. If you do not see a linear pattern in the factors, try doubling the original terms in the sequence and factor those numbers. When you find the pattern and write out the nth term, don't forget to divide by two.

6. If multiplying by two doesn't work, try multiplying by three (or another logical number; if you have fifths, multiply by 5).

Example: 8, 15, 24, 35, 48, 63, 80

Factors:

1, 8 1, 15 1, 24 1, 35 1, 48 1, 63 1, 80

2, 4 3, 5 2, 12 5, 7 2, 24 3, 21 2, 40

3, 8 3, 16 7, 9 4, 20

4, 6 4, 12 5, 16

6, 8 8, 10

If we use the factors (2, 4) (3, 5) (4, 6) (5, 7) (6, 8) (7, 9) and (8, 10) we can see the first number in each set of factors creates a nice linear sequence: 2, 3, 4, 5, 6, 7, 8 the formula for the nth term for this sequence is n + 1. The second set of numbers also creates a nice linear pattern: 4, 5, 6, 7, 8, 9, 10. The formula for the nth term for these numbers is n + 3. (See the instructions for finding the nth term of a linear sequence if you need a reminder.) So the formula for this quadratic sequence is (n + 1)(n + 3). Note we are just multiplying the formulas for the factors to get the original sequence, just as we multiply the factors themselves to get the actual numbers.

Example 2: -8, 11, 42, 85, 140, 207, 286

Factors:

1, -8 1, 11 1, 42 1, 85 1, 140 1, 207 1, 286

2, -4 2, 21 5, 17 2, 70 3, 69 2, 143

-1, 8 3, 14 4, 35 9, 23 11, 26

-2, 4 6, 7 5, 28

7, 20

If we use the set of factors (-1, 8) (1, 11) (3, 14) (5, 17) (7, 20) (9, 23) and (11, 26) the first set of numbers in these factors go up by two each time and the other numbers go up by 3 each time so they are both linear. The formula for the first term is (2n - 3) and the formula for the second term is (3n + 5), so the formula for the quadratic sequence is (2n - 3)(3n + 5).

Example 3: -4, -3, -6/5, 7/5, 24/5, 9, 14

You'll notice right away that we have fifths in this example. The logical thing to do would be to multiply by 5 so that we have whole numbers, then factor. So we will find the factors of: -20, -15, -6, 7, 24, 45, 70

1, -20 1, -15 1, -6 1, 7 1, 24 1, 45 1, 70

2, -10 3, -5 2, -3 2, 12 3, 15 2, 35

4, -5 -3, 5 -2, 3 3, 8 5, 9 5, 14

-4, 5 -1, 15 -1, 6 4, 6 7, 10

-2, 10

-1, 20

If we use the sets (-5, 4) (-3, 5) (-1, 6) (1, 7) (3, 8) (5, 9) (7, 10) the first set of numbers increases by two each time and the second set of numbers increases by 1. The formula for the first set of numbers is (2n - 7) and the formula for the second set is (n + 3). The formula for the quadratic sequence then is ((2n - 7)(n + 3)) /5. It is very important you don't forget to divide by 5 at the end because we multiplied by 5 before finding factors!

Cubic We know that a cubic will have the form n^3 + n^2 + n with possible coefficients before each of these and perhaps adding a constant at the end: an^3 + bn^2 + cn + d. We will use this knowledge and the fact that if you put in 1 for n you get the first term in the sequence, if you put 2 in for n you get the second term in the sequence, etc. Cubics are best explained with an example. The following is a cubic sequence: 10, 34, 88, 184, 334.

So we know that if we put in 1 for n we will get 10 for an answer:

(1) a + b + c + d = 10 (note I put in 1 for n)

We also know that if we put 2 in for n, we will get 34:

(2) 8a + 4b + 2c + d = 34.

We will continue to do this because we know if we have 4 unknowns, we need four equations:

(3) 27a + 9b + 3c + d = 88

(4) 64a + 16b +4c +d = 184

Now I will use elimination to solve these equations. Notice that I have numbered each equation so that you can see which equations I am subtracting and given them new numbers so that I can refer to that line again later:

(5) = (2) - (1): 7a + 3b + c = 24

(6) = (3) - (2): 19a + 5b + c = 54

(7) = (4) - (3): 37a + 7b +c = 96

(8) = (6) - (5): 12a + 2b = 30

(9) = (7) - (6): 18a + 2b = 42

(10) = (9) - (8): 6a = 12

Now we can solve for a: a = 2.

Now we will use substitution to solve for b, c and d:

Substituting 2 in for a in equation (8) we find b = 3.

Substituting 2 in for a and 3 in for b in equation (5) we find c = 1.

Substituting 2 in for a, 3 in for b and 1 in for c in equation (1) we find d = 4.

The formula for this cubic sequence is 2n^3 + 3n^2 + n + 4.

Series vs. Sequences A sequence is simply a list of numbers which follow a pattern. A series, is when you have a list of numbers following a pattern and you are adding the numbers all together! An arithmetic sequence is a sequence with the difference between two consecutive terms constant. We have called these linear sequences in the past, either name is acceptable. The difference between terms is called the common difference (d).

To find the nth term of an arithmetic sequence you take the first term and add (n-1) times the common difference. t1 + (n-1)d A geometric sequence is a sequence with the ratio between two consecutive terms constant. You multiply by the same number each time. This ratio (or number you multiply by) is called the common ratio (r).

To find the nth term of a geometric sequence, you take the first term and multiply it by the common ratio to the (n-1) power. t1*r^(n-1) 10

Σ 2i

i=1 The symbol Σ is the capital Greek letter sigma and mathematically it means to add all the numbers up. The number under the Σ tells you what term to start with. In this case we are to start with the first term. The number above the Σ tells you when to stop. In this example, we are to stop when we reach the tenth term. The numbers to the right of the Σ tell us the actual equation we are using. So we would substitue 1 in for i and get 2, then we substitute 2 in for i and get 4, then we substitute 3 in for i and get 6 and we keep doing this until we substitute 10 in for i and get 20. Then we add all these numbers together: 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110 To find the sum of an arithmetic series, you use the following formula: S=(n/2)(a1+an), where S is the sum, n is the number of terms, a1 is the first term and an is the last term.

For example: 2+4+6+8+10+12+14+16+18+20

S=(5)(2+20) There are ten terms so n=10 the first term (a1) is 2 and the last term (an) is 20. Therefore S=110.

To find the sum of a geometric series, you use this formula: S=(a1(1-r^n))/(1-r), where S is the sum, a1 is the first term, r is the common ratio and n is the number of terms.

For example: 2+6+18+54+162+486

S=(2(1-3^6))/(1-3) = (2(1-729))/(1-3) = (2(-728))/(-2) = 728 Visit my website for a power point of review questions and answers!