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Lecture Comments (3)

1 answer

Last reply by: Professor Selhorst-Jones
Fri May 9, 2014 6:11 PM

Post by abendra naidoo on May 8, 2014

hello,
I came across this problem in my reading and I am stumped as to how the answer can have 2 terms with a plus in between:Find the exact sum of the first 14terms in the geom. sequence sqrt.2,2,2sqrt2,4�
The answer given is 254+127sqrt2.
Using formula Ssubn=Asub1(1-r^n)/1-r --- I can't see how you end up with a plus sign in the answer?
Thanks.

Introduction to Series

  • Given some sequence a1, a2, a3, a4, …, a series is the sum of the terms in the sequence (or a portion of them).
  • If the sequence is infinite, we call it an infinite series. It adds all of the terms together:
    a1 + a2 + a3 + a4 + …
    If the sequence is not infinite or we only wish to add up a finite number of its terms, we call it a finite series. Adding the first n terms of the sequence together is called the nth partial sum:
    a1 + a2 + a3 + …+ an
  • To compactly describe sums, we use sigma notation (sometimes also called summation notation).


     
    ai
    • ai is the thing being summed. Terms from the sequence given by ai are added together. This might be a sequence, but is more often an algebraic expression.
    • i is the index of summation. It increases by 1 for each "step" of summation. The index can be any symbol, but i is common.
    • Above, i=1 is the first value used for the index. This is the Lower Limit of Summation.
    • Above, n is the last value used for the index this is the Upper Limit of Summation.
    Sigma notation can be really confusing the first few times you use it. Check out the video to see an example of how it works and some picture diagrams.
  • To show an infinite series (one where the terms keep adding forever), we put `∞' on top of the sigma. This shows that the series has no upper limit and instead continues on forever.


     
    ai    =  a1 + a2 + a3 + a4 + a5 + …
  • Sometimes it is useful to reindex a series (or sequence). We might have a series that has the index begin at one value, but we want it to start at another. However, we can't just change the number for the lower limit, because that would affect the whole series. This means we have to think about how to alter every part of the sigma notation so we can get the index we want without changing the value of the series. There are two main ways of doing this:
    • Expand: Write the sigma notation out in its expanded form, then look for how we could rewrite the pattern with our chosen starting index in mind.
    • Substitution: How does the old index relate to the new index that we want? Set up that relation, then use substitution to find the new general term and upper limit.
  • Sums have various properties that we can occasionally use to our advantage. Below, let ai and bi be sequences, let c be a constant.
    • i=1nc = c·n
    • i(c ·ai) = c ·∑i ai
    • i(ai + bi) = ∑i ai + ∑i bi

Introduction to Series

Given the sequence below, find the 4th partial sum. Assuming the pattern continues, find the 6th partial sum as well.
25,   18,   11,   4,   …
  • The nth partial sum of a sequence is the summation of the first n terms of the sequence. For example, the 3rd partial sum of a sequence would add up the first, second, and third terms.
  • Finding the 4th partial sum is quite easy: we already have the first four terms, so just add them all together:
    25 + 18 + 11 + 4    =     58
  • Finding the 6th partial sum takes a little more work: we don't know the rest of the sequence yet, so we need to write out more terms. Looking at the sequence, we see the pattern is to subtract by 7 for every step. Thus our next two terms can be figured out as below:
    4 − 7 = −3     ⇒     −3 − 7 = −10
    Thus, continuing the sequence to the sixth term, we have
    25,   18,   11,   4,   −3,   −10,   …
  • Now that we know the first six terms, it's easy to find the 6th partial sum:
    25 + 18 + 11 + 4 + (−3) + (−10)     =     45
4th partial sum: 58,        6th partial sum: 45
Compute the value of the below sum:
(21+1)  + (22+2)  + … + (26+6)
  • To understand how to sum this up, we need to realize what the ellipsis (`...') means. In math, an ellipsis between terms or after terms means that the pattern continues as expected.
  • For this problem, the pattern is clear: the nth term is (2n + n). Since we don't have very many terms to add, it's easiest to just write out each term in the sum to compute the sum's value.
  • Write out each term following the pattern:
    (21+1)  + (22+2)  + (23+3)  + (24+4)  + (25+5)  + (26+6)
    From there, just work through the arithmetic:
    (2+1)  + (4+2)  + (8+3)  + (16+4)  + (32+5)  + (64+6)

    (3)  + (6)  + (11)  + (20)  + (37)  + (70)

    147
147
Expand the below sigma notation into a series of terms added together, but do not add the terms together. Just leave it in a format where the pattern of the terms is easy to see.
5

i=2 
( i2 + 3i)
  • Sigma notation (also called summation notation) allows us to compactly write a series. If you are unfamiliar with it, make sure to watch the video lesson: it is explained carefully and with helpful diagrams to show how it works.
  • Remember, the variable below the sigma (Σ) shows what changes, and the number the variable is equal to shows where to start. The number at the top is where we end, and the expression to the right gives how to create the terms.
  • We start by plugging in i=2 for the first term, then i=3, then i=4, then we finish at i=5 because that's the value at the top of the sigma.
    5

    i=2 
    ( i2 + 3i)     =     (22 + 3·2)  +  (32 + 3·3)  +  (42 + 3·4)  +  (52 + 3·5)
    Since the problem told us to leave it in a form where the pattern of the terms is obvious, we're done.
(22 + 3·2)  +  (32 + 3·3)  +  (42 + 3·4)  +  (52 + 3·5)
Compute the value of the below series:
8

k=3 
(2k−7)
  • Sigma notation (also called summation notation) allows us to compactly write a series. If you are unfamiliar with it, make sure to watch the video lesson: it is explained carefully and with helpful diagrams to show how it works.
  • Remember, the variable below the sigma (Σ) shows what changes, and the number the variable is equal to shows where to start. The number at the top is where we end, and the expression to the right gives how to create the terms.
  • For this problem, there aren't too many terms, so it's easiest to just expand out all the terms from the sigma notation then add them up by hand. [If the series had many terms, it would be quite time-consuming. Luckily, there are formulas for various types of series, as we will see in later lessons. If you have to add more than 10 or 20 terms together, you're probably better off trying to find a formula.] We start by plugging in k=3 and stepping up until we hit k=8:
    8

    k=3 
    (2k−7) =     (2·3−7)  + (2·4−7)  + (2·5−7)  + (2·6−7)  + (2·7−7)  + (2·8−7)
    From there, just work through the arithmetic:
    (6−7)  + (8−7)  + (10−7)  + (12−7)  + (14−7)  + (16−7)

    (−1)  + (1)  + (3)  + (5)  + (7)  + (9)

    24
24
Expand the below sigma notation into a series of terms added together, using an ellipsis (`...') to show the pattern continuing. Do not compute the sum, just leave it in a format where the pattern of the terms is easy to see.
47

v=−3 
(v3+8)
  • Sigma notation (also called summation notation) allows us to compactly write a series. If you are unfamiliar with it, make sure to watch the video lesson: it is explained carefully and with helpful diagrams to show how it works.
  • Remember, the variable below the sigma (Σ) shows what changes, and the number the variable is equal to shows where to start. The number at the top is where we end, and the expression to the right gives how to create the terms.
  • It would take a very long time and a lot of space to write out the entirety of the series. However, using an ellipsis, we can just write enough terms to convey the pattern, along with showing where it starts and stops. Write out the first three terms added together (or however many is needed to clearly see the pattern), then put an ellipsis, then the final term to show where it stops. For this problem, we start at v=−3, then stop at v=47.
    47

    v=−3 
    (v3+8)     =     ( (−3)3 + 8 )  +  ( (−2)3 + 8 )  +  ( (−1)3 + 8 )  +  … +  ( (47)3 + 8 )
    Since the problem told us to leave it in a form where the pattern of the terms is obvious, we don't want to simplify anything (since that would make the pattern less clear).
( (−3)3 + 8 )  +  ( (−2)3 + 8 )  +  ( (−1)3 + 8 )  +  … +  ( (47)3 + 8 )
Use sigma notation to represent the below sum. Write it in a such way so that the lower limit of summation (the number it "starts" at) is 1.
7  +  9  +  11  +  13  +  … +  33
  • Sigma notation (also called summation notation) allows us to compactly write a series. If you are unfamiliar with it, make sure to watch the video lesson: it is explained carefully and with helpful diagrams to show how it works. Remember, the variable below the sigma (Σ) shows what changes, and the number the variable is equal to is the lower limit of summation (where it "starts"). The number at the top is where we end, and the expression to the right gives how to create the terms.
  • Notice that writing something in sigma notation means we need a formula for any given term in the series. This is the same as figuring out the nth term for a sequence, so consider the terms of the series as a sequence:
    7,   9,   11,   13,   …
    With this in mind, it's fairly easy to see that the pattern adds 2 each time and starts at 7. We can express this as
    an = 2n+5
    [If you are unfamiliar with sequences and/or finding formulas for the nth term of a sequence, check out the previous lesson, Introduction to Sequences. It will be very helpful for understanding the more difficult problems about series.]
  • We set it up so that a1 = 7, so our starting index of i=1 in the sigma notation is already set. We just need to figure out what the upper limit of summation needs to be (where it "ends"):
    ?

    i=1 
    (2i+5)
    Looking at the series in the problem, we see that it must end at 33, so we can find what n value will create 33 from the an formula:
    33 = 2n+5     ⇒     28 = 2n    ⇒     14=n
    Thus our series must end on 14:
    14

    i=1 
    (2i+5)
    Finally, always make sure to check your series. It's easy to make a mistake, so expand your series to see the first couple terms and the last term to confirm it against the original series.
    14

    i=1 
    (2i+5)     =     (2·1 + 5)  + (2·2 +5)  + … + (2·14 + 5)     =     7  +  9  +  … +  33    
i=114 (2i+5)
Use sigma notation to represent the below sum. Write it in a such way so that the lower limit of summation (the number it "starts" at) is 1.
(−7)  +  (−4)  +  1  +  8  +  17  +  28  +  41  +  … +  392
  • Sigma notation (also called summation notation) allows us to compactly write a series. If you are unfamiliar with it, make sure to watch the video lesson: it is explained carefully and with helpful diagrams to show how it works. Remember, the variable below the sigma (Σ) shows what changes, and the number the variable is equal to is the lower limit of summation (where it "starts"). The number at the top is where we end, and the expression to the right gives how to create the terms.
  • Notice that writing something in sigma notation means we need a formula for any given term in the series. This is the same as figuring out the nth term for a sequence, so consider the terms of the series as a sequence:
    −7,   −4,   1,   8,   17,   28,   41,   …
    This is a fairly tough pattern to crack, but after realizing it's not based on a basic operation, we will eventually compare it to the pattern of n2:
    n2     ⇒     1,   4,   9,   16,   25,   36,   49,   …
    This is a challenging pattern to spot, but we can see that the sequence from the series matches up to the n2 sequence, it's just 8 below every term. Thus we can give the sequence for the series as
    an = n2 − 8
    [If you are unfamiliar with sequences and/or finding formulas for the nth term of a sequence, check out the previous lesson, Introduction to Sequences. It will be very helpful for understanding the more difficult problems about series.]
  • We set it up so that a1 = −7, so our starting index of i=1 in the sigma notation is already set. We just need to figure out what the upper limit of summation needs to be (where it "ends"):
    ?

    i=1 
    (i2−8)
    Looking at the series in the problem, we see that it must end at 392, so we can find what n value will create 392 from the an formula:
    392 = n2−8     ⇒     400 = n2    ⇒     ±20 = n
    Since the upper limit must be higher than the lower limit, we throw away the −20 and see that it ends on 20:
    20

    i=1 
    (i2−8)
    Finally, always make sure to check your series. It's easy to make a mistake, so expand your series to see the first couple terms and the last term to confirm it against the original series.
    20

    i=1 
    (i2−8)     =     (12 − 8)  + (22−8)  + … + (202 − 8)     =     (−7)  +  (−4)  +  … + 392    
i=120 (i2−8)
Use sigma notation to represent the below sum. Write it in a such way so that the lower limit of summation (the number it "starts" at) is 1. [Notice it is an infinite series.]
1

1
 +  1

4
 +  1

9
 +  1

16
 +  1

25
 + …
  • Sigma notation (also called summation notation) allows us to compactly write a series. If you are unfamiliar with it, make sure to watch the video lesson: it is explained carefully and with helpful diagrams to show how it works. Remember, the variable below the sigma (Σ) shows what changes, and the number the variable is equal to is the lower limit of summation (where it "starts"). Normally, the number at the top is where we end, but in this case the series never ends, so we will have an ∞ sign. Finally, the expression to the right gives how to create the terms.
  • Notice that writing something in sigma notation means we need a formula for any given term in the series. This is the same as figuring out the nth term for a sequence, so consider the terms of the series as a sequence:
    1

    1
    ,    1

    4
    ,    1

    9
    ,    1

    16
    ,    1

    25
    ,   …
    With this in mind, it's fairly easy to see that the pattern is just a fraction with 1 on top and a denominator given by n2. We can express this as
    an = 1

    n2
    [If you are unfamiliar with sequences and/or finding formulas for the nth term of a sequence, check out the previous lesson, Introduction to Sequences. It will be very helpful for understanding the more difficult problems about series.]
  • We set it up so that a1 = [1/1], so our starting index of i=1 in the sigma notation is already set. Normally we would need to figure out what the upper limit of summation needs to be (where it "ends"):
    ?

    i=1 
    1

    i2
    However, notice that this series is infinite: it continues the pattern forever. We show this by putting an ∞ on top of the sigma:


    i=1 
    1

    i2
    Finally, always make sure to check your series. It's easy to make a mistake, so expand your series to confirm it against the original series.


    i=1 
    1

    i2
        =     1

    12
     +  1

    22
     +  1

    32
     + …    =     1

    1
     +  1

    4
     +  1

    9
     + …   
i=1 [1/(i2)]
Reindex the below series so it starts at the index i=1.
100

j=15 
(4j+7)
  • Reindexing a series means changing the way the lower limit, upper limit, and expression are written, but still getting the same terms. For a very simple example, the series on the left has been re-indexed into the series on the right, but they still give the exact same terms and result:
    11

    k=10 
    k            =            2

    l = 1 
    (l+9)
    Make sure to check out the video lesson for more information and a much more comprehensive explanation.
  • There are two different ways to approach reindexing. First, we'll look at it from the point of view of expanding. Expand the sigma notation so that you can see the terms it is built on:
    100

    j=15 
    (4j+7)     =     (4·15 +7)  + (4·16 + 7)  + … + (4 ·100 + 7)

                  = 67  + 71  + … + 407
    Now that we see it in its expanded form, we can approach putting it in sigma notation like we normally would. We see that it has a pattern of adding 4 each term and starts at 67, so the general term is
    an = 4n + 63
    Next, we need to know when it stops, so we solve for the appropriate n to create an=407:
    407 = 4n+63     ⇒     344 = 4n     ⇒     86 = n
    Thus the sigma series will end at the 86th term. Putting this all together, we get
    86

    i=1 
    (4i+63)
  • Alternatively, we could approach reindexing by substitution. Notice that we want to have a starting index of i=1, but we currently have j=15. However, they must indicate the same first term, so we see that
    if i=1 and j=15,    then    j = 14+i.
    With this in mind, we can now substitute for the bottom index and the index variable in the expression. For now, we'll leave the top index as a `?' since we haven't figured out what it must be yet. Plug in:
    100

    j=15 
    (4j+7)     ⇒     ?

    14+i=15 
    (4(14+i)+7)     =     ?

    i=1 
    (56+4i+7)     =     ?

    i=1 
    (4i+63)
    Finally, we need to figure out the top index. Notice that in terms of j, the final value was 100. Thus we can plug in j=100 to the conversion formula to find the matching i-value:
    j=100     ⇒     100 = 14+i     ⇒     86 = i
    Therefore the top index is 86 in the new i-format, which completes the sum:
    86

    i=1 
    (4i+63)
    [As an alternative to find the top index, we could also see that the original top index was 85 above the starting index (100−15 = 85), so the new top index must still be 85 above the new starting index (1+85=86).]
  • Whatever method you use to find the reindexed series, make sure to check that it still contains the same terms as the original series. It's easy to make a mistake, so double-check your answer:
    100

    j=15 
    (4j+7)    ⇒     67  + 71  + … + 407     ⇐     86

    i=1 
    (4i+63)    
i=186 (4i+63)
Using the functions below, calculate f(−5) and g(3).
f(x) = 4

i=0 
xi           
           g(t) = t

k=1 
4k
  • Plugging into a function is just like we're used to, even if it's mixed up with a series. Simply replace the variable the function uses as input with the given number, then work out the arithmetic.
  • For f(−5), since the function is f(x), we plug in x=−5:
    f(−5) = 4

    i=0 
    (−5)i
    From there, just calculate the series like usual:
    4

    i=0 
    (−5)i     =     (−5)0  + (−5)1  + (−5)2  + (−5)3  + (−5)4

                         =     1  + (−5)  + 25  + (−125)  + 625     =     521
    Thus f(−5) = 521.
  • For g(3), since the function is g(t), we plug in t=3:
    g(3) = 3

    k=1 
    4k
    From there, just calculate the series like usual:
    3

    k=1 
    4k     =     4·1  + 4·2  + 4·3     =     4+8 + 12     =     24
    Thus g(3) = 24.
f(−5)=521,        g(3) = 24

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Introduction to 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 0:00
  • Introduction 0:06
  • Definition: Series 1:20
  • Why We Need Notation 2:48
  • Simga Notation (AKA Summation Notation) 4:44
    • Thing Being Summed
    • Index of Summation
    • Lower Limit of Summation
    • Upper Limit of Summation
  • Sigma Notation, Example 7:36
  • Sigma Notation for Infinite Series 9:08
  • How to Reindex 10:58
    • How to Reindex, Expanding
    • How to Reindex, Substitution
  • Properties of Sums 19:42
  • Example 1 23:46
  • Example 2 25:34
  • Example 3 27:12
  • Example 4 29:54
  • Example 5 32:06
  • Example 6 37:16

Transcription: Introduction to Series

Hi--welcome back to Educator.com.0000

Today, we are going to talk about an introduction to series.0002

In the previous lesson, we introduced the idea of a sequence--that is, an ordered list of numbers.0005

With this idea in mind, we can now discuss the concept of a series, summing up the terms of a sequence.0011

Now, at first, this might seem a little bit series.0018

It is just addition, and we have been doing addition since kindergarten; so why do we need to talk about addition in a special way?0020

But consider how long it would take to add up 100 terms by hand.0025

Could you imagine how much time it would take up if you had to add up 100 different things from a sequence?0029

What if it were a thousand terms, or ten thousand terms?0033

The study of series can make these seemingly colossal summation tasks, having to add up huge numbers of numbers, really, really easy.0037

We can just turn this stuff into being some trivially easy task once we figure out how to talk about series in a deep way.0044

Being able to easily add up lots of numbers has many applications.0051

You will see it in science, engineering, economics, computer programming, advanced math, and many other fields.0054

All of these things benefit greatly from the study of series,0061

from being able to talk about adding up a whole bunch of numbers in an easy way,0064

where we can talk about it in compact notation--all of these things are really important to a variety of fields.0068

So, this is a great thing to study; let's start by defining a series.0073

Given some sequence a1, a2, a3, a4...0078

a series is the sum of the terms in the sequence; it is just adding up the terms,0083

or maybe a portion of them--maybe not the entire sequence.0088

But it is just adding up terms from the sequence.0091

If the sequence is infinite, and we are adding up all of the terms from the sequence, we call it an infinite series.0093

It adds all of the terms together; so I am saying that it keeps going forever.0099

It is a1 + a2 + a3 + a4...going on forever...0104

so plus a5, a6, a7, a8...forever and ever and ever.0110

Since our sequence was infinite, we are just saying to keep adding them forever and ever and ever.0114

That is an infinite series.0119

On the other hand, we can talk about a sequence that is not infinite--or if we only wish to add up a finite number of its terms.0121

We aren't going to add them forever; we are just adding up a finite portion of them--just a number where we can count how many are there.0129

We call this a finite series; adding the first n terms of the sequence together is called the nth partial sum.0136

If we add a1 + a2 + a3, up until a1,0144

but notice: it just stop there--we don't go any farther past an--then that is the nth partial sum,0149

because we are adding in all of the terms, 1, 2, 3, up until we get to n; so it is the nth partial sum,0154

because it is only a part of our infinite sequence.0160

Consider if we had a sequence to find by the general term an = n + 2n over 3n.0164

That is 1 + 2 over 3, 2 + 22 over 32, 3 + 23 over 33,0170

4 + 24 over 34...what if we wanted to talk about its thirtieth partial sum?0176

In talking about its thirtieth partial sum, we will see very quickly why we need a special form of notation for series.0182

Why is writing this stuff out by hand going to be a real pain?0189

If we wanted to write this out, we would have to first show the pattern that occurs here.0192

We have this pattern that created each one of these various terms.0197

We would have to have that show up in our summation; otherwise, we wouldn't be able to realize what pattern is going on.0202

So, we are going to at least need the first three terms--at least.0208

Then, we could use an ellipsis, that ... , to say that the pattern continues on in this manner.0213

We can not have to write a massive amount, because we can use the ellipsis to say that the pattern keeps going.0218

But we are still going to have to show its stop at the thirtieth term.0223

At a very minimum, we are going to have to write out the first three terms, the last term, and an ellipsis in the middle.0226

We would have to write out (1 + 2)/3 + (2 + 22)/32 + (3 + 23)/33 + ...0232

+ (30 + 230)/330; that is the thirtieth partial sum, like we were talking about.0240

But that is a lot of stuff to write out; that is a lot of writing for a fairly simple series.0246

This isn't even that complicated; but if we had to write this on multiple lines, as we worked through a problem--0250

if we had to keep talking about this over and over as we did steps in a problem--your wrist is going to hurt after doing one of these problems.0256

And you are going to have to do a bunch of problems about summation.0261

This is why we need some sort of notation: we need notation to make it easier to compactly describe a series.0263

We don't have to write out this really, really long thing every time we want to talk about some summation from a sequence.0271

We need some easy way to be able to do it in a short way.0278

Enter sigma notation: to compactly describe sums, we use something called sigma notation.0281

It is also sometimes called summation notation.0287

Why is it called sigma notation?--because it uses the uppercase Greek letter sigma.0290

Sigma is this right here; if you are drawing it by hand (this is a computer-drawn picture of sigma,0295

made by a computer typesetting program), I would recommend writing it like this,0302

where you have these little vertical hooks on either end; and then it is just kind of a capital M on its side.0312

Just write it out like that; and that is great.0319

If you are feeling lazy, you can end up just sort of writing it like that, as well.0323

But it helps to write it like this, so that we can clearly see what it is.0326

But if you are doing a lot of problems that way, don't worry if it ends up getting not quite absolutely perfect each time,0329

because you will be able to recognize it from other things that are about to be seen.0335

All right, let's see how we use sigma notation.0338

Let's look at the anatomy of a series in sigma notation.0342

The first thing to look at is the thing being summed: the terms from the sequence given by ai, this thing to the right of the sigma, are added together.0346

This is what will be added together over and over with each step as we add up:0356

the first thing, and then the second thing, and then the third thing, and so on.0360

It might be a sequence, but it is much more often going to be an algebraic expression--something like 3 times i plus 10 or 2 to the i or i!,0362

something that we could plug in a number and be able to churn out a value.0373

You will often see algebraic expressions; but once in a while, you will see a sequence like ai.0376

Next, the index of summation; i is the index--it increases by 1 for each step of summation.0382

We do the first thing, and then we add the second thing in the next step;0391

and then we add the next thing in the next step; and we add the next thing in the next step, and so on, and so forth.0394

So, every time we step forward, the index will click up by 1.0399

We start at some value, and then we go to the value one above that, then the value one above that, then the value one above that, and so on.0403

So, every step, it will increase by 1.0409

The index can be any symbol; but i is very common, and that is what we will end up using for the most part in this course.0412

So, that is the thing that ends up changing; and notice how i will generally occur over here in the series, as well.0419

That is saying that this is the thing that changes; this is what will change with each step--this little thing right here.0425

The lower limit of summation: this is the first value used for the index.0432

This is where the series starts; in this case, since i = 1, our first value would be plugging in a 1 for the i.0436

The upper limit of summation: this is the last value used for the index, where the series ends.0444

We would step up until we eventually got to some value n.0450

All right, let's see it in action: we have this pictorial summary of how it works.0454

This is a great picture; so check against this, if you get used confused in a later point; this is a great slide right here.0458

Let's see it in action: we have sigma...it's i = 3 for the starting index, and then 7 is the upper limit, and we have 2i - 1 as the actual expression.0464

The very first thing: we have the i right here, and we start at i = 3.0475

So, the first value that we plug in for our i is 2 times 3, because our lower limit of summation was that 3.0479

So, 2 times 3 minus 1 is the first thing.0486

Then, we go on to our next step: plus...we increase our index by 1.0489

Our index started at 3, so it goes from 3...plus 1...to 4; so 2 times 4, minus 1.0493

Then, we go on to our next step; it will be 2 times...we were at 4 in the last step, so it increases to 5; so we are at 4 + 1 is 5.0500

2 times 5 minus 1, plus...our index increases by 1 again, so 2 times 6 minus 1, plus...0508

and now, finally, we have just hit our final, upper limit.0516

So, at this point, we stop; this will be the last one, once we click up to this final upper limit for how high our index goes to.0520

That is the last one in the series; so it is 2 times 7 minus 1.0528

And at this point, we could simplify it out if we wanted to.0533

We could get a value from this; but that is also a good way to see how it expands.0535

And that is all we are looking for right now--seeing how it expands.0539

But if you wanted to, you could simplify each one of these.0541

And you could also combine them and be able to get a value for that series.0543

So far, we have only seen how to use sigma notation for finite sums--that is to say, partial sums, not sums that are going on forever.0548

Since they have all had some upper limit of summation, something on top of this sigma, they have all had to eventually stop.0555

They have had some top limit to how far they step out.0561

If we want to show an infinite series, one where the series' terms keep adding forever and ever,0564

we simply put an infinity symbol on top of the sigma.0570

This shows that the series has no upper limit, and instead continues on forever.0573

So, if we had a sigma with i = 1 on the bottom, our index is i, and our lower limit is 1,0577

and our upper limit is infinity, that says really "just keep going."0583

It is not that we actually get to infinity and stop; we can't ever get to infinity.0586

Infinity is just the idea of going on forever; so we start at 1, and then on to 2, and then on to 3, and then on to 4, and then on to 5, and then on to 6...0589

And it just keeps going forever and ever.0598

So we would have...since it is ai, we would have a1 first, because it was a 1 for our lower limit,0600

then a2, and then + a3, and then + a4, and then + a5...0605

and it will just keep going on forever, because it is an infinite series that we are working with.0609

If we wanted to see in specific, here is an example that uses specific numbers.0614

We have some series with i = 1 on the bottom; it is an infinite series, because of the infinity on top.0618

We have 1/2 to the i; so our first i is 1; that is 1/2 (is our first term, one-half), plus 1/2...0623

our next step with the i will be now at a 2, so 1/2 to the 2, 1/4...and then our next step would be i at 3;0633

so 1/2 to the 3; that is 8; at our next step, i will be at 4, so that is 1/2 to the 4, so 1/16.0641

And then, i will be at 5, so it is 1/2 to the fifth, so 1/32; and that pattern will just keep going.0648

We will keep adding them on forever and ever.0653

Sometimes it is useful to re-index a series (or a sequence, sometimes).0657

We might have a series that has the index begin at one value, but we want it to start at another value.0662

For example, we might want a transformation like this one here, where we have, in our original sequence,0667

i = 7 as our starting index, and it has some upper limit, and it has some expression that we are actually working with.0672

And we want to transform it to k = 1; we want our starting index to be k = 1.0678

And we are going to, of course, also need some upper limit and something here.0684

Now, of course, we can't just change the number of the lower limit.0688

If we just went and changed the lower limit, that would affect the whole series.0691

We have to pay attention to how the thing being summed, and our upper limit, end up making the transformation, as well.0696

They are going to change over the process; they are not going to be the same thing on the left side and the right side.0703

They are going to end up becoming different things; otherwise we would have altered the whole thing by starting...0710

we start at one starting place, and then we just change the starting place, and we don't do anything else;0715

well, if you start in a different starting place, you are going to have different values coming out.0718

So, since we want to end up having the same value come out of our two ways of talking about it,0721

we have to alter every part of the sigma notation, so that we can get the index we want without changing the value that comes out of it.0725

You are probably asking why we care; if we have it written in one way, why not leave it that way?0734

Well, there are a bunch of reasons to do it.0738

For example, if you are working on a proof, it can sometimes help to re-index it.0740

But specifically to this course, a lot of formulas that we end up working with will be given in the form of something with an i = 1 on the bottom.0743

So, since a lot of our formulas will have i = 1 on the bottom, we will have to start at this index of something = 1,0752

some symbol = 1, to be able to work with some of the formulas we have.0758

A lot of the formulas that you end up seeing are in this format.0762

So, it is really helpful to be able to re-index, so we get to the format that we already have a formula for,0764

as opposed to having to figure out an entirely new formula for some new, different index.0768

So, how do we actually re-index?0773

The most essential way is to expand the sigma notation into a written-out series,0775

then see how we could rewrite the pattern with our chosen starting index in mind.0779

So, for example, if we want to go from i = 7 to k = 1, we can just start by...0783

a series is just a shorthand way of writing out something + something + something...+ something.0791

And even that is a shorthand way of writing out every single term.0797

So, we can expand it into that format; if it is i = 7 first, then we would have 3(7) - 16;0800

the next thing would be + 3 times 1 step up to 8, minus 16; plus...it would continue in this form.0807

And 3 times...our final upper limit is 22; so 3(22) - 16; great.0815

If we want, we can figure out how to write this sigma over here, with just that in mind.0822

But we can also simplify things to see if we can see another easier-to-see pattern.0827

So, 3(7) would give us 21, minus 16, plus...3 times 8...16, 24...minus 16 + ... 3(22) is 6, minus 16...0831

so, 21 - 16 is 5, plus 24 - 16 (that is 8), plus...plus 66 - 16 (that is 50).0845

We look at this, and we might realize, "Oh, what is doing is going up by 3 each time."0854

And that makes sense, since we started at 3i.0858

It is going up at 3 each time; if that is the case, and we want to get this k = 1, then that means0860

that we know that here, we are at k = 1; here we are at k = 2; here we are at k =...0865

we will actually leave that as a question mark right now, because we don't quite know yet.0870

We will talk about that in just a moment--what number we are at; we will see why.0873

k = 1; if it is + 3 each time, we are going to want some 3k + some number.0877

So, if we are at 5 here, then that would be 3k + 2, because 1 for k...3 times 1 plus 2 does get us 5, so that checks out.0882

Here, with 3k + 2, it works, as well; so 3k + 2 is here; we plug in 2; 3(2) is 6, plus 2 is 8; that checks out, as well.0889

So, it looks like it is going to be some 3k + 2 that will be here; we will have 3k + 2 for that part right there.0897

Now, what is the upper limit of summation going to end up being?0905

Well, if we are at 3k + 2, and it equals 50, we have to figure out what value would end up coming out here.0908

So, k = ?; we could solve for this: 3k = 48; k...divided by 3 now on both sides...we get 16.0915

So, we know that our upper limit is going to have to be 16.0922

Another way of doing it is to notice what the difference is between our upper limit and our lower limit.0925

In our original, we had 22 as our upper and 7 as our lower; so, 22 - 7 means 15.0930

Oops, I'm sorry; not 15...oh, it is 15; 22 - 7 comes out to be 15.0936

Over here, we know that we are going to end up having that difference of 15, as well.0944

Since we start at k = 1, 1 plus that same 15 (because we are going to have to have the same number of steps,0949

however we phrase it) comes out to be 16, which is the same thing we figured out that it has to be to work there.0958

With that in mind, we now know that our lower limit with our index is going to be k = 1,0965

because that is what we wanted to figure out in the first place.0972

Our top, our upper limit of summation, will be 16; we have figured out two different ways that that has to be the case,0974

because 16 - 1 is 15, which is the same as 22 - 17; or alternately, we can solve for it0980

using the formation that we came up with for that general term.0985

And then, 3k + 2 is what is actually going in that is being added on each term.0988

And so, that is one way of doing it.0994

We worked this out by expanding first; expanding is a very specific way to see it.0995

It is a great thing to do if you get confused by the problem that you are working on.1000

There is another way to do this, though; we can also re-index by thinking in terms of substitution.1004

How do our old index and the new index we are creating relate to each other?1008

We can think in terms of substitution: consider, once again, if we wanted to convert from i = 7, with all of the same thing, to k = 1.1012

Well, since we have i = 7 and k = 1 both at their starting places, how is i related to k?1020

Well, i is the same thing as k + 6; 1 + 6 gives us 7.1026

So, we now have this relationship where i is equal to k + 6.1031

With this in mind, we can substitute to figure out what the upper limit is, and then to figure out what the general term is.1036

To figure out the upper limit, we know that i = 22 is what we are going to plug in, since that is its upper limit.1043

And then, we are going to...over here, we know i = k + 6, so we plug in 22 for i: we have 22 = k + 6.1051

Subtract 6 on both sides, and we get 16 = k for its upper limit.1062

So, at this point, we can now write that our sigma is going to end up being...upper limit of 16;1072

the lower starting index is at k = 1; now, what is it going to end up being?1080

Well, here it was 3i - 16; so we have 3; what is our i in terms of k?1085

Well, i is k + 6; and then, we continue on with the rest of it...minus 16.1093

We have substituted out the i here for the k + 6 here, into what i was in our initial version for the sigma notation.1100

So, at this point, we can expand this and work this out, and simplify it if we want.1111

We could also just leave it as it is, and it would be just fine.1114

The upper limit and lower limit and our index--they will all stay the same throughout.1117

We work this out: 3 times k plus 6 gets 3k +...3 times 6 is 18; minus 16; and so, we get 16 for our upper limit; k = 1 for our index and lower limit.1121

3k + 18 simplifies to 3k + 2, which is exactly what we had the first time when we did this through expanding it.1135

So, substitution and expanding both work the same way.1146

Substitution is probably a little bit faster and easier, but it is a little bit more complicated to see what is going on, to really understand.1148

The important thing is to figure out how your i...whatever your initial symbol and the new symbol that you are changing to are...1153

how they are related to each other; and you probably start with the lower limit,1159

and then ask what your upper limit will have to be, to have that same relationship between old index and new index.1162

And then, what is your expression going to end up being, if you just plug in your connection between the two indexes?1168

All right, working with series, there are various properties that we can occasionally use to our advantage.1177

Let's look at some properties of sums: below, let ai and bi be sequences.1181

They are things that are liable to change; but c will simply always be a constant.1187

Our first one is that the sum from i = 1 to n of c is equal to c times n.1191

Remember: c is just a constant.1197

Why is this the case? Well, remember: if we had the sum of i = 1 to n of c,1198

well, c doesn't change any time, because it is a constant; it isn't affected by the index.1206

So, it is just going to be c, plus c, plus c...plus c, that many times.1210

If we have that, how many times did it show up?1216

Well, we went from i = 1 up until n, so 1, 2, 3, 4...we count up to n; we have a total of n terms in there.1219

If c adds to itself n times, then that is just c times n.1227

So, that is equal to c times n; and that is where we get that property for how sums work.1233

If it is just a constant being added through a series, we can just multiply it by the number of terms in that series.1239

Next, we have the summation of some summation...and this will end up working for any summation limit;1246

the only thing that we have to care about is the index.1253

You can start with any lower limit and any upper limit, including an infinite sum; it is still going to end up working out the same.1255

So, c times ai...we can do this where we pull out the constant, and we bring it out to the front.1261

And it is c times the summation of ai.1267

So, why is this the case? Well, notice: our first term would be c times a1.1270

Plus...our next term would be c times a2; plus...our next term would be c times a3.1276

And it is going to continue on in this pattern.1283

It might end; it might not end; we will just leave it as dots there.1285

But notice: we have a c on each one of our terms.1288

So, if we want, since we have a c here, a c here, a c here, and that thing is going to end up continuing1291

(there is going to be a c on each one of these, because we see it is c times ai),1295

we can pull out all of the c's; we pull them out, so it is c times a1 + a2 + a3 + ....1299

So, it is c multiplied against that entire series; it is not going to have any difference in how we do it.1310

So, this entire series, here to here--we can think of it as c times a new series of just the ai's changing.1315

And that is how we end up getting the same thing here.1323

The final idea: if we have addition, ai + bi, well, we can end up breaking that into two separate series, added together.1327

And once again, this would end up working with any limit; so that is why we don't have a lower limit and don't have an upper limit--just an index.1335

It is because this will work with any limits whatsoever, even an infinite series.1340

We could write this...if it is series of i, ai + bi, well, that is going to be a1 + b1,1346

plus a2 + b2, plus a3 + b3...1353

And it is just going to continue on in that format.1360

Well, order of addition doesn't matter; a1 + b1 + a2 + b21363

is the same thing as a1 + a2, then plus b1 + b2.1368

So, what we do is order all of our a1's first; we will have all of the a1's in our series show up first.1371

And then, we will add on our b1's: b1 + b2 + b3...1381

And that will also go on the same as it would have on its own.1388

So, at that point, what we have done is just re-ordered how we are looking at the series, since we have expanded it,1393

and now we have re-ordered the way we are looking at it.1397

But we haven't changed anything; so at this point, we can just pull it into two separate series, i...1399

so, we will have ai here, plus...and then the bi.1404

There is the a portion of the series and the b portion of the series.1410

So, we can either have them intermixed together, or we can spread them apart and work on each of them on its own.1413

And that is how we end up having this final property.1419

All right, we are ready for some examples.1422

The first example here: Given the sequence below, find the third partial sum.1424

If we are going to find the third partial sum, that is as easy as just adding up the first three terms.1429

That is 0, 3, and 8; 0 + 3 + 8; we end up getting 11; it's done; it was easy.1433

The next one: If we are looking for the seventh partial sum...well, so far, we only have 0, 3, 8, 15, 24...1443

so we are going to have to figure out what the pattern will continue on to.1448

Our first step is to figure out how this pattern ends up working.1452

We look at this through addition: 0 to 3, 3 to 8, 8 to 15...well, the addition is changing each time.1455

So, we could think in terms of some recursive relationship, but that is not really going to make it easier.1460

Multiplication? Multiplication doesn't really work, either.1464

But we look at this for a while, and we might realize that this looks kind of like 1, 4, 9, 16, 25...1466

0, 3, 8, 15, 24...that is just - 1 on each one of the terms.1476

So, we can think of this as being, in general, n2 - 1.1480

If that is the case, then we can figure out what the next terms are.1486

If it is n2 - 1, the next term to follow is going to end up being 36 - 1, because we are at 5 here; 52 - 1 is 24.1489

So, the next will be 62 - 1; that is going to get us 35.1497

The next one will be 72 - 1, 49 - 1, or 48.1502

At this point, the seventh partial sum will be adding up these first seven terms.1507

So, 0 + 3 + 8 + 15 + 24 + 35 + 48...we toss that all into a calculator, and we end up getting 133; there we go.1511

The next one: Expand the below sigma notation into a series of terms; add them together using an ellipsis.1530

We break it apart into the thing where we are not using sigma notation.1536

We are just writing it as one term after another, with some pattern occurring.1539

So, what would our very first term be?1542

Our very first term would start at i = 3; so we plug in a 3 for our i first; that would be 3! over 73.1544

Plus...our next term would end up being...we click up one, from 3 to 4; that will be 4! over 74.1554

Plus...our next one: we click up another one to 5! over 75, plus..., plus...we are going to end up finishing here at 15.1564

We could also start with something before 15; for example, we could write out 14!,1574

because that would be in the expansion: 714, plus 15! over 715.1579

There we go; we have managed to write the whole thing out, using ellipsis.1589

We have expanded the sigma notation into a series of terms.1591

And we don't have to necessarily write out all of these.1596

We could probably get away with not writing it.1599

We could certainly get away with not writing this one right here, the 14!/714.1601

And we might even be able to get away with not writing the 5!/75.1605

It helps us see the pattern, but it is not absolutely necessary.1609

But by including more terms, sometimes the reason we want to expand things is so we can manipulate how the terms work.1612

So, sometimes it is useful to include more things at the start and the end, so that we can end up seeing how things are interacting.1618

It will sometimes allow things to work out better.1625

And we will see that sometimes, when we are working in proofs.1627

Example 3: Condense the sum below into a series expressed using sigma notation.1631

Notice: it is an infinite series.1635

The first thing: let's notice how these connect to each other; how do we get 1/2 - 1/4 + 1/8 - 1/16?1637

Well, we can break this into a sequence, first, that just has a series going on.1643

So, let's look at this underlying sequence; how do we get from 1/2 to -1/4? Well, we multiply by -1/2.1647

How do we get from -1/4 to positive 1/8? We multiply by -1/2.1653

How do we get the next one? We multiply by -1/2.1658

So, with this in mind, how can we create a general term an?1661

That is going to be an =...since we are multiplying by -1/2 each time, it will be -1/2 to the...1665

is it going to be to the n? Well, here we are going to end up starting with something else, actually.1672

It is n - 1, because here is n = 1; we want to not have anything at the beginning, which we could write as times 1/2... so -1/2 to the n - 1, times 1/2.1677

Alternatively, another equivalent way to write this out: we could write this as -1 to the n - 1 over 2 to the n - 1, times 1/2.1687

The 2's in the denominator will end up compacting together.1699

And we could also write this as -1 to the n - 1, over 2 to the n.1701

Both of these are just fine ways to write it.1710

We will end up getting the same thing, whether we write it using this general term, or we end up writing it with this general term.1711

We will end up getting the same series.1718

What is our first term? We decided to set it at some n = 1; let's use an index of i,1720

just because it is what we are used to, although we could use n.1724

Sometimes, you might have a little bit of confusion, since we are used to talking about nth terms with n.1727

So, it might be weird to use an index of n; but you will often see it used, as well.1731

I like i, so I am going to use i.1734

i = 1, because that is our first location; what do we go up to?1737

Well, it is an infinite series, because it goes on forever, and we are never told that it stops.1740

So, we use an infinity symbol on top; and then, let's do this one first.1745

We could write this as -1/2 to the n - 1, times 1/2; and that whole thing has the series applied to it.1749

Or equivalently, this would also be equal to the series--the same upper limit of going on forever, and the same starting index.1759

We could also write this as -1 to the n - 1 over 2 to the n.1770

Either way we end up working it out, both of these are just fine.1775

They are going to give us the exact same answer; they are just two different ways to write it out.1780

Depending on the specific problem, it might be useful to write it one way or the other.1783

But any teacher, if this was just the question, should accept both of these.1787

All right, calculate the value of sigma with a lower limit of 0 and an upper limit of 4.1791

These are indexed where the expression is v! - 5v.1796

The first thing we do is plug in v = 0 for our first 0.1800

Our very first term is going to be 0! - 5(0).1805

Our index is v, so that is the thing being swapped out.1810

Plus...we step up our index to the next level, 0 up to 1 now; so 1! - 5(1).1813

Plus...the next thing, stepping up from 1, is 2; so 2! - 5(2).1822

Plus...the next thing, from 2 to 3: 3! - 5(3); and then, plus...3 to 4...4! - 5(4).1829

Finally, at this point, we notice that that is our upper limit, so we stop.1841

We step from our lower limit, 0, 1, 2, 3, 4...our upper limit, so that is where we stop.1845

And we do each of the steps in between and add them all together.1851

So now, it is just a matter of simplifying to actually get the value.1853

What is 0!? Remember: 0! is simply defined to be equal to 1.1856

All of the other numbers, factorial, are that number, times each of the positive integer numbers underneath it.1861

But 0! is just simply defined to be 1.1868

So, 0! gets us 1; minus...5 times 0 is 0, plus 1! is just 1 times itself; so 1 minus...5 times 1 is 5, plus 2!...1872

2 times 1 is 2, minus 5(2) is 10, plus 3! is 3 times 2 times 1, so it is 6; minus 5(3) is 15;1883

plus 4! is 4 times 3 times 2 times 1, 24; minus 5(4) is 20.1893

We work this out; we have 1 here, plus -4, plus -8, plus -9, plus +4.1901

We see that we have a +4 here and a -4 here; they cancel out.1912

1 + -9 gets us -8, so we have -8 and -8, or -16, once we simplify the whole thing out.1917

Great; the fifth example: Re-index the series below so that it starts at the index k = 1.1924

We are looking to swap the index i = 5 out for k = 1, but have the exact same value come out of the series.1930

Our sigma notation will change; the upper limit will end up changing, and the expression here will end up changing to some extent,1937

so that we can achieve this without affecting the value of the series.1943

We talked about two different ways to do this when we went through the lesson.1947

The first way we will approach this is through expanding it.1951

We will start by doing it through expanding it; and then later, we will look at doing it through substitution.1957

There are two alternative ways; whichever one makes more sense to you, that is the one that you probably want to end up using for your own work.1964

All right, we want to start by expanding this.1969

If we are going to expand i = 5, 37 - i, that is going to be...our first thing would be 37 - 5, so 32.1972

Plus...then the next would be 6 for our i; so 37 - 6 would be 31,1981

plus...then at 7, we will have 30, plus...plus...and then finally,1990

when we get up to an upper limit of 20, 7 - 20 is -13, so it will be 3-13.1997

OK, that is what we see that we end up getting out of this: 32 + 31 + 30...+ 3-13.2004

So, we see that the top steps down each time.2011

We want it to figure out how we can end up showing this.2015

We start with k = 1 for this first slot; use k = 2, k = 3...and here is k = we-don't-know-what-yet.2017

We can actually figure out that it is going to have to be...20 - 5 is 15, so 1, our starting lower limit here, plus 15 would be 16.2026

We know it will end up having it come out to 16; but let's work through it the other way.2034

So, what we have here...it looks like we could write this as 33 - k.2038

33 - 1 would get us 32; the next one, 33 - 2, would get us 31; that works out.2045

At k = 3 on our third term, we would have 33 - 3; it would be 30; that ends up working out.2054

So, if we are going to have 3-13 = 33 - k, we are now going to know that -13 has to be equal to 3 - k.2059

So, we have k = 16 as our upper limit, as well; great.2069

That ends up making sense; we can also figure it out, once again, from:2076

our starting upper limit was 20; our starting lower limit was 5.2079

So, that means that there is a difference of 15; so if we have k = 1, then 1 + 15 has to come out to be 16, as well.2083

So, that is going to be our upper limit; there are two ways of looking at it.2093

Either way, just make sure that you are careful with this sort of thing.2095

It is easy to get confused the first few times you work with it.2097

So now, we know what our general term is; it is 33 - k.2100

We know what our upper limit is, and we know what our index and starting term is, and lower limit.2103

So, we can write the whole thing out: we have 16 as our upper limit; k = lower limit of 1; 3 to the 3 - k; great.2109

So, that is our answer; and that is doing it through expanding as our method.2120

Alternatively, we could do this through substitution.2125

We could have also set this up by noticing that we have i = 5 here, and we want to start at k = 1.2128

So, at the first place, we have i = 5; and then, at our second place, we have k = 1.2133

So, if we have i = 5 here and k = 1 here, then we can see that i is equal to k + 4.2139

So, if i is equal to k + 4, in general (this is just going to hold true in general), then what about our upper limit?2146

Our upper limit: it is going to have to end up being the case that when i is equal to 20...what will our k end up being?2154

We will have 20 = k + 4, which means 16 is equal to k for the upper limit.2160

16 = k for our upper limit; now we just plug in through substitution.2167

So, 16 is our upper limit; we just figured that out; k = 1 is our index and lower limit.2172

So, it is going to be the same thing we started with, 37 - i, so 3 to the 7 minus...2178

but we are not going to use i; we are going to use i = k + 4.2186

So, we are going to swap out this i here for k + 4.2190

We substitute that in, and we have k + 4.2194

We can now work this out and simplify it, since it is the upper limit and lower limit and index...2199

none of those will change; we are just simplifying what is on the inside.2205

3 to the 7 - (k + 4), so minus k and minus 4...7 - 4 gets us 3, and minus k gets us 3 - k in the whole.2208

So, we end up getting the exact same thing, either way we end up approaching this: through substitution or through expanding.2218

We can end up re-indexing the series, so we can have it change at a different starting index.2224

They both work fine; whichever one makes more sense to you, that is the one I would recommend using.2229

All right, the final example: Calculate the value of the series below; use summation properties to make the math less tedious.2233

We could just do this by hand; we could say, "All right, we have 3 times 1 minus 5, plus...plus...all the way up to 3 times 15 minus 5."2240

OK, and then we could calculate what 3 times 1 minus 5 is, and what 3 times 2 minus 5 is, and what 3 times 3 minus 5 is.2252

And we could do this by hand or through the calculator; but that is kind of a pain.2258

That is going to be a lot of writing; that is going to be a lot of calculation.2262

Luckily, there are some clever ways to use the summation properties that we learned early to make this at least...2265

not easy...not fast (it will be easy), but less tedious, so we have to do less writing.2270

How can we do this? Well, first, remember: we can split based on addition.2276

We can see this as 3i + -5; and we can split this into i = 1; the limits to our summation will never end up changing,2280

but we can split this into a 3i +...the same sum over here...15i = 1 on -5; cool.2290

We also were allowed to pull out constants; we have this constant of a 3 here; we have this constant of a -1 here, effectively.2299

We can pull those out front; we can now write this as 3 times the summation of 15, i = 1 of i, plus...2306

Now, I am sorry...plus -1, so let's just write that as minus...summation...15...i = positive 1 (I got confused by the negative) on 5; great.2319

So, 3 times...well, there are no cool properties that we have learned yet, although we will learn in the very next lesson,2330

how to easily sum this one up; but we would have 1 plus 2 plus 3 plus...plus 15.2338

And over here, we will have minus...well, we could do 5 + 5 + 5...but it is just going to show up 15 times.2345

We have i = 1 to 15; we talked about this before.2351

Since a constant is just showing up 15 times, that is going to be 15 times the constant.2354

So, it will be 15 (the number of terms that are there) times the constant that is showing up over and over, so 15 times 5.2359

We can now work out with a calculator 1 + 2 + 3 +...up until 15.2366

That ends up working out to 120; so we have 3(120) - 15(5) is 75; 3 times 120 is 360, minus 75; that comes out to equal 285, and there is our answer.2372

We still had to do a little bit of writing this out.2389

The slowest part here is going to be adding by hand: 1 + 2 + 3 +...to 15, or by using a calculator.2391

But either way, that is much better than having to multiply numbers, and then subtract 5, and then add that, each time, over and over.2399

So, we can use these summation properties to split things up into ways that make them easier to work with.2406

All right, cool; in the next lesson, we will end up looking at our first specific kind of sequence and series, arithmetic.2411

And then later on, we will work with geometric, which will let us apply these ideas about series into one specific thing,2416

where we can actually start creating some formulas to make things really easy and fast.2421

All right, we will see you at Educator.com later--goodbye!2424