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Algebra 2 Geometric Series

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Section 11: Sequences and Series: Lecture 4 | 22:43 min

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Lecture Comments (9)

	1 answer Last reply by: Dr Carleen Eaton Thu Oct 9, 2014 11:38 PM Post by Peter Spicer on August 22, 2014 in example 2, on the divisor you put the r term equal to 8 when its supposed to be 4.
	0 answers Post by julius mogyorossy on April 29, 2014 Solving equations reminds me of being in combat.
	1 answer Last reply by: Dr Carleen Eaton Sun Oct 21, 2012 7:46 PM Post by Chonglin Xu on October 13, 2012 Hey, Dr. Eaton, I thought I found your error at 5:53. The correct thing is "3+9+27+81+243+729", but NOT "3+9+27+81+729+2187". Did you do the double check of Sigma notation example?
	1 answer Last reply by: Dr Carleen Eaton Thu Jul 19, 2012 5:17 PM Post by Daran Daneshjou on July 17, 2012 How do I remove a comment?
	1 answer Last reply by: Daran Daneshjou Tue Jul 17, 2012 3:03 PM Post by Daran Daneshjou on July 17, 2012 Why is there an extra 162 in the numerator of example #1?

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Geometric Series

There are 2 formulas for the sum. Use one if you are given n. Use the other if you are given a_n. In both cases, you will be given the values of r and a₁.
To find a particular term of a series, say the 4^th term, you first need to find the first term using the formula for S_n. Then multiply the first term by r³, using the formula for the nth term of a series, a_n = a₁r^n-1.

Geometric Series

Find S_n for the geometric series with
a₁ = − 4;r = − 5;n = 7

To find the sum, use the equation S_n = [(a₁(1 − rⁿ))/(1 − r)]
S_n = [(a₁(1 − rⁿ))/(1 − r)]
S₇ = [( − 4(1 − ( − 5)⁷))/(1 − ( − 5))] =
S₇ = [( − 4(1 − ( − 78125))/(1 + 5)]
S₇ = [( − 4(1 + 78125))/6]

S₇ = [( − 4(78126))/6] = [( − 4(78126))/6] = [( − 312504)/6] = − 52084

Find S_n for the geometric series with
a₁ = − 3;r = 3;n = 9

To find the sum, use the equation S_n = [(a₁(1 − rⁿ))/(1 − r)]
S_n = [(a₁(1 − rⁿ))/(1 − r)]
S₉ = [( − 3(1 − 3⁹))/(1 − (3))] =
S₉ = [( − 3(1 − 19683))/( − 2)]

S₉ = [( − 3( − 19682))/( − 2)] = [59046/( − 2)] = − 29523

Find S_n for the geometric series with
a₁ = − 1;r = 4;n = 8

To find the sum, use the equation S_n = [(a₁(1 − rⁿ))/(1 − r)]
S_n = [(a₁(1 − rⁿ))/(1 − r)]
S₈ = [( − 1(1 − 4⁸))/(1 − (4))] =
S₈ = [( − 1(1 − 65536))/( − 3)]

S₈ = [( − 1( − 65535))/( − 3)] = [65535/( − 3)] = − 21845

Find the sum of the first 6 terms of the geometric series:
− 2 − 6 − 18...

To find the sum, use the equation S_n = [(a₁(1 − rⁿ))/(1 − r)], but first, look for the common ratio r
r = [(a₂)/(a₁)] =
r = [(a₂)/(a₁)] = [( − 6)/( − 2)] = 3
Find the sum, using r, n = 6, and a₁
S_n = [(a₁(1 − rⁿ))/(1 − r)]
S₆ = [( − 2(1 − 3⁶))/(1 − (3))] =
S₆ = [( − 2(1 − 729))/( − 2)]

S₆ = [( − 2( − 728))/( − 2)] = [1456/( − 2)] = − 728

Find the sum of the first 8 terms of the geometric series:
2 + 6 + 18 + 54...

To find the sum, use the equation S_n = [(a₁(1 − rⁿ))/(1 − r)], but first, look for the common ratio r
r = [(a₂)/(a₁)] =
r = [(a₂)/(a₁)] = [6/2] = 3
Find the sum, using r, n = 8, and a₁
S_n = [(a₁(1 − rⁿ))/(1 − r)]
S₈ = [(2(1 − 3⁸))/(1 − (3))] =
S₈ = [(2(1 − 6561))/( − 2)]

S₈ = [(2( − 6560))/( − 2)] = [( − 13120)/( − 2)] = 6560

Find the sum of the first 9 terms of the geometric series:
3 − 9 + 27 − 81...

To find the sum, use the equation S_n = [(a₁(1 − rⁿ))/(1 − r)], but first, look for the common ratio r
r = [(a₂)/(a₁)] =
r = [(a₂)/(a₁)] = [( − 9)/3] = − 3
Find the sum, using r, n = 8, and a₁
S_n = [(a₁(1 − rⁿ))/(1 − r)]
S₉ = [(3(1 − ( − 3)⁹))/(1 − ( − 3))] =
S₉ = [(3(1 − ( − 19683)))/4]

S₉ = [3(19684)/4] = [59052/4] = 14763

Find S_n for the geometric series with
a₁ = 3;a_n = 192;r = 2

To find the sum, use the equation S_n = [(a₁ − a_nr)/(1 − r)]
S_n = [(a₁ − a_nr)/(1 − r)]
S_n = [(3 − 192*2)/(1 − 2)]

S_n = [(3 − 384)/(1 − 2)] = [( − 381)/( − 1)] = 381

Find S_n for the geometric series with
a₁ = 2;a_n = 31250;r = − 5

To find the sum, use the equation S_n = [(a₁ − a_nr)/(1 − r)]
S_n = [(a₁ − a_nr)/(1 − r)]
S_n = [(2 − 31250*( − 5))/(1 − ( − 5))]

S_n = [(2 + 156250)/(1 + 5)] = [156252/6] = 26042

Find S_n for the geometric series with
a₁ = 2;a_n = − 256;r = − 2

To find the sum, use the equation S_n = [(a₁ − a_nr)/(1 − r)]
S_n = [(a₁ − a_nr)/(1 − r)]
S_n = [(2 − ( − 256)*( − 2))/(1 − ( − 2))]

S_n = [(2 − (512))/(1 + 2)] = [( − 510)/3] = − 170

Find S_n for the geometric series with
a₁ = 2;a_n = − 524288;r = − 4

To find the sum, use the equation S_n = [(a₁ − a_nr)/(1 − r)]
S_n = [(a₁ − a_nr)/(1 − r)]
S_n = [(2 − ( − 524288)*( − 4))/(1 − ( − 4))]

S_n = [(2 − 2097152)/(1 + 4)] = [( − 2097150)/5] = − 419430

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Answer

Geometric Series

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

What are Geometric Series?

Sum of Geometric Series

Example: Sum of Geometric Series

Sigma Notation

Another Sum Formula

Example: n Unknown

Specific Terms

Example 1: Sum of Geometric Series

Example 2: Sum of 8 Terms

Example 3: Sum of Geometric Series

Example 4: First Term

Intro 0:00
What are Geometric Series? 0:11

List of Numbers
Example: Geometric Series

Sum of Geometric Series 2:16

Example: Sum of Geometric Series

Sigma Notation 4:21

Lower Index, Upper Index
Example: Sigma Notation

Another Sum Formula 6:08

Example: n Unknown

Specific Terms 7:41

Sum Formula
Example: Specific Term

Example 1: Sum of Geometric Series 10:02
Example 2: Sum of 8 Terms 14:15
Example 3: Sum of Geometric Series 18:23
Example 4: First Term 20:16

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Geometric Series

Welcome to Educator.com.0000

Today we are going to talk about geometric series.0002

In the previous lesson, I introduced the concept of geometric sequences; so this continues on with that knowledge.0004

So, what are geometric series? A geometric series is the sum of the terms in a geometric sequence.0012

Again, make sure that you have geometric sequences learned, that you understand that well, before going on to geometric series.0017

But just briefly, recall that a geometric sequence is a list of numbers.0025

And what is unique about this list is that you find one term by multiplying the previous term by a number r, which is the common ratio.0030

For example, here the common ratio is 2: so I multiply 8 times 2 to get 16, times 2 is 32, times 2 is 64.0042

You can always find that common ratio by taking a term and dividing it by the previous term.0053

This is a geometric sequence: today we are going to move on to talk about geometric series.0062

And a geometric series is the sum of the terms; so from this, we could get a geometric series, 8 + 16 + 32 + 64.0072

This is the geometric series: first term + second term + third term, and on and on, until we get to that last term.0083

Now, this is a finite series, but you could also have an infinite series, where it just continues on indefinitely.0100

So, what we want to find, often, is the sum of a particular number of terms in the series.0112

Now, I could look up here and say, "OK, I want to find the first three terms: that is 8 + 16 + 32."0121

And then, I could just add that up and figure out what it is.0128

But that is going to get really cumbersome to add manually; so we have a formula for the sum of a geometric series.0132

The sum of the first n terms of a geometric series is given by this formula.0139

And we have the limitation that r does not equal 1, because if r equaled 1, if I had, say, 3,0143

and I just multiplied it by the common ratio of 1, I would just get 3 again and again and again, and it wouldn't really change.0151

So, the limitation is that the common ratio cannot equal 1.0158

Looking at an example, 6 + 18 + 54 + 162, let's say I wanted to find the sum of this entire series--the sum of all the four terms.0162

I could use this formula: my first term here is 6--what is my common ratio?0174

Well, I can say 18/6 is 3; so the common ratio is 3.0179

Therefore, the sum of these four terms would be 6, times (1 - 3ⁿ) (n = 4 in this case),0186

divided by (1 - 3); this is going to give me 6(1 - 3⁴).0197

Well, recall that 3 times 3 is 9, times 3 is 27, times 3 is 81; so this is 81, divided by 1 - 3.0208

So, the sum is 6 times...1 - 81 would give me -80, divided by...1 - 3 is -2.0219

Let's go up here to continue on: 6 times -80 is going to give me -480, because 6 times 8 is 48; add a 0;0231

divided by -2; the negatives cancel out; 480/2 is 240.0241

So, I was able to find this sum by using a formula, rather than just adding each number.0248

And the formula requires that I know the common ratio, the first term, and n.0255

Sigma notation: as with arithmetic series, we can also use sigma notation as a concise way to express a geometric series.0261

Again, the Greek letter Σ means sum; and the variable that we are going to use is called the index.0270

So here, I have a lower index; and then the upper index tells me how high to go.0278

So, here I have n going from 1 to 5; and then I am going to have the formula for the general term0286

written right here, so I know how to find each term in this geometric series.0292

Looking at an example that is specific: as I said, they often use the letter i for the index in sigma notation, so I am going to use i.0297

i going from 1 to 6...and the formula to find each term is going to be 3 raised to the i power.0308

Therefore, I can find this series; it is going to be 3 raised to the first power, plus 3...0316

I started out with 1, and I inserted that here; next I am going to go to 2.0328

Then, I am going to go to 3, then 4, 5, and 6.0333

And then, you could, of course, figure this out; this is 3 + 9 + 27 + 81 + 729 + 2187.0343

So, this notation means this; and again, this is just a different way of writing a geometric series.0355

But the concepts that we have discussed remain the same.0365

All right, we learned one sum formula: and there is a second formula.0368

This second formula is very valuable when you know the first and last terms, and you know the common ratio, but you don't know the number of terms.0374

If n is not known, use this formula.0382

For example, maybe I have a series, and I know that the first term is 128; that the last term is 4; and that the common ratio is 1/2.0388

But I don't know n--n is not known--and I am asked to find the sum.0400

I can do that using this formula: the sum is going to be the first term, minus this last term, times r, divided by (1 - r).0410

which is going to be equal to 128 minus...4 times 1/2 is 2...divided by...1 minus 1/2 is just 1/2.0422

This is going to be equal to 126 divided by 1/2, and that is the same as 126 multiplied by the reciprocal of 1/2, which is 2.0432

And that is 252; so the sum is going to be 252, and I was able to find that because I knew the first term; I knew the last term; and I knew the common ratio.0442

And I had a formula that did not require me to know n.0457

All right, the formula for s_n can be used to find a specific term in the series.0462

And a very important term is the first term: so we often use s_n to find the first term in a series.0468

So, looking at one of the sum formulas that we just discussed, that would be the first term, times (1 - rⁿ), divided by (1 - r).0477

Let's say that you are given that the sum is 62.0489

And you are also given that the common ratio is 2, and that the number of terms is actually 5.0496

And you want to find the first term; you can do that with this formula.0504

I know that 62 equals the first term, times 1 - 2, raised to the n power (here, n is 5), times 1 - 2.0509

So, 62 = a₁...2 to the fifth...2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32; so that is 32.0521

1 - 32, divided by -1...therefore, this equals...rewriting this as 62 =...the first term...1 - 32 would give me -31,0533

divided by -1; if you just pull the negative out front, then those are going to end up canceling; -31a/-1 is just going to give you a positive.0551

So, I am going to rewrite this as 62 = 31 times the first term, because this is just a -1 down here, and I can rewrite this much more simply.0570

OK, dividing both sides by 31 gives me that the first term equals 2.0585

So, I was able to find the first term by being given the sum and the common ratio and the number of terms.0590

Frequently, you will use one of your sum formulas to find the first term in a geometric series.0597

All right, on to Example 1: Find the sum, s_n, for the geometric series with a first term of 162,0603

with a common ratio of 1/3, and with a number of terms equal to 6.0613

All right, the formula for the sum: a₁(1 - rⁿ)...and since I know n, I can use this formula,0619

rather than the other one...divided by (1 - r); and I want to find s_n.0632

s_n equals the first term, which is 162, times (1 - 1/3⁶), all of that divided by (1 - 1/3).0639

Therefore, the sum equals...162 times 1, minus 162 times 1/3 to the sixth power, divided by...1 - 1/3 is going to leave me with 2/3 in the denominator.0655

Now, this looks like it might be complicated to work with; but there is a shortcut.0672

If you think of...I have my 162; I can think of 162 as 2 times 81, and 81 is a multiple of 3, so you might be able to see where I am going with this.0680

2 times 81...as I said, that is a multiple of 3; 3 times 3 is 9, times 3 is 27, times 3 is 81.0698

Therefore, I could rewrite this as 162 - 2...let's move this out of the way a bit...times 3 to the fourth power, times 1/3 to the sixth power.0708

I can use rules governing how I work with powers, and say, "OK, if this gives me 3⁴/3⁶, 162 -2 times...0725

if I look at this as 3⁴/3⁶, they have the same base; if I take 6 - 4, this is going to give me 1/3²."0742

So, this is 2 times (1/3)², divided by 2/3--it is much easier to work with now.0762

This is simply going to be 162 minus 2 times 1/9, or 2/9; all divided by 2/3.0772

Coming up here to finish this out: the sum, therefore, equals 162 - 2/9; that is going to give me 161 and 7/9, all divided by 2/3.0785

Recall that, then, dividing by 2/3 is the same as multiplying by 3/2.0799

That is not a very pretty answer, but you can simplify this, calculate it out, use a calculator...but this does give the sum.0810

So again, I was given the first term, the common ratio, and the number of terms.0818

I can use this formula, since I have the number of terms.0822

This looked like it was going to be very messy to work with: 162 times (1/3)⁶.0825

But by recognizing that I could break 162 into 2 times a multiple of 3, I was able to get this into a base with 3, 3⁴.0830

Dividing 3⁴ by 3⁶ canceled out, and I got (1/3)²;0842

and that simplified things a lot to give me this answer that I have in the upper left.0848

Find the sum of the first 8 terms of the geometric series.0855

So, here we are asked to find the sum of the first 8 terms of this series, and that would be s₈.0859

Thinking about which formula I am going to use: I know n; since I am looking for the first 8 terms, I know n.0867

I can also easily find r, because I can just take 1 divided by 1/4; 1 divided by 1/4 is the same as 1 times 4, or 4.0877

So, I have r; I have n; the other thing is the first term, and I have that.0888

With these three pieces of information, I know that I can use the formula0895

that the sum equals the first term, times (1 - rⁿ), divided by (1 - r).0898

So, the first 8 terms...that is going to be 1/16, times 1 minus...r is 4...raised to the n power, where n is 8, divided by (1 - 8).0904

OK, let's look at what I have: I have 1/16 times 1, minus 1/16 times 4⁸.0932

Actually, this is not 1 - n; it is 1 - r; so let's go ahead and change that to a 4.0955

All right, and then I have 1/16 times 4⁸; as with the previous problem,0960

you might recognize that there is something to make this a lot easier than taking 4 to the eighth power and dividing it by 16.0965

You are going to recognize that you can rewrite this 1/16 as (1/4)².0971

So, 1/4 times 1/4 would give me 1/6; that times 4⁸ allows me to do some canceling to make things simpler.0985

1 - 4 just gives me -3; let's go up here and work this part out.0998

This is (1/4)² times 4⁸; this is the same as taking 4⁸ and dividing it by 4².1003

Using my rules governing exponents, 4⁸/4²...if I want to divide, and I have like bases,1015

I just subtract the exponents; so this is going to give me 4 raised to the sixth power.1028

So, knowing your rules of exponents is important to solve when you are working with this type of problem.1032

Therefore, I am going to get 1/16 - (1/4)² times 4⁸1038

(that is the same as 4⁶, so I am going to go ahead and rewrite it that way) divided by -3.1044

At this point, you are going to have to do a lot of multiplying, or else use your calculator to determine that 4⁶ is actually 4096.1052

And we are dividing by -3; so the sum is 1/16 - 4096; that comes out to -4095 and 15/16, all divided by -3.1059

And a negative and a negative will give me a positive, so that would give me 4095 and 15/16, divided by 3.1077

You can work this out with your calculator and see that this sum is approximately equal to 1365.3.1086

This does give us our answer; but we can estimate to put it in a neater decimal form.1094

Finding the sum: this time, we are given the first term; we are given the last term; and we are given the common ratio.1104

But we don't know the number of terms--we don't know what n is.1113

But that is OK, because we have that other formula that allows us to find the sum of a geometric series when we know the first term;1117

we know the last term; and we know the common ratio (that second formula that we worked with).1125

All right, looking for the sum: I have the first term; this is 4 - a_n (which is 8748), times the common ratio of 3, all of that divided by 1 - 3.1140

That is going to give me that the sum equals 4 minus...multiply this out, or use your calculator, to get 26244, divided by 1 - 3 (gives me -2).1159

Coming up here to the second column: the sum is going to be equal to 4 - 26244, or -26240, divided by -2.1174

A negative divided by a negative gives me a positive; and 26240 is even--I divide it by 2, and I get a nice whole number, 13120.1192

So, the sum of this geometric series is 13120; and I found that using the formula1201

that only requires that I know the first term and the last term, and the common ratio.1207

I didn't have to know the number of terms in the series, and I could still find the sum of the terms.1211

Example 4: We are asked to find the first term.1217

As I mentioned, it is important to have the first term frequently when you are working with these series.1219

And you can use the sum formulas to find that first term.1224

I am going to look at the information I have: I have the sum; I have the common ratio; and I have the number; but I don't have the last term.1227

So, I use the formula that involves the number of terms, not the last term.1233

And what I am looking for is the first term; therefore, the sum is 1020, equals the first term,1244

times 1 - r (is 2), and it is 2 raised to the eighth power, divided by 1 - 2.1254

And it is 2 raised to the eighth power, divided by (1 - 2); this is going to give me 1020 equals the first term, times 1 minus...1258

if you go through your powers of 2, you will find that 2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32.1270

So, 2⁵ - 32; then, we are going to get 2⁶ is 64; 2⁷ is 128; and then, 2⁸ is 256.1283

That is 1 - 256, divided by 1 - 2 (is -1); therefore, 1020 = a₁...1 - 256 is -255; divided by -1.1297

Therefore, 1020 = -a₁ times 255, divided by -1.1316

Well, a negative and a negative gives me a positive, so that is 1020 = a₁ times 255.1325

I just take 1020 and divide by 255; so I have divided both sides by 255.1335

And I am going to get that the first term equals 4.1340

I was asked to find the first term, and I found that using the sum formula requiring the first term, the common ratio, and the number of terms.1345

And I determined that the first term in this geometric series is 4.1353

That finishes up this lesson on geometric series on Educator.com; thank you for visiting!1358

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Geometric Series

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro