Get ready for Precalculus
The finite geometric series formula is a(1-rⁿ)/(1-r). In this video, Sal gives a pretty neat justification as to why the formula works.
Want to join the conversation?
- Is it possible to find n by using a formula, as it is with arithmetic series?(5 votes)
- The video is actually about geometric series, however it is useful some knowledge regarding arithmetic series.
It will depend on the exact question.
How many number are there from 0-150?
Ans: 150 - 0 + 1 = 151
There is the plus one because we need to include 0.
How many numbers are there in the given sequence:
0, 2, 4, ...., 20
If we divide by 2 we get:
0, 1, 2, ..., 10:
Ans: 10 - 0 + 1 = 11 numbers
How many numbers are there in the sequence:
7, 9, 11, ..., 21
Subtract by 7 to get:
0, 2, 4,..., 14
Divide by 2:
0, 1, 2, ..., 7
Therefore the amount of numbers is 7-0+1 = 8(1 vote)
ok... I'm so confused! s of n? a's and r's? I have no idea what's going on, help? I would appreciate it! thanks! :D(3 votes)
- So the majority of that video is the explanation of how the formula is derived. But this is the formula, explained:
Sₙ = a(1-rⁿ)/1-r
Sₙ = The sum of the geometric series. (If the n confuses you, it's simply for notation. You don't have to plug anything in, it's just to show and provide emphasis of the series.
a = First term of the series
r = the common ratio
n (exponent) = number of terms.
As an example:
What is the sum of the 4,16,64,256?
The common ratio is 4, as 4 x 4 is 16, 16*4 = 64, and so on.
The first term is 4, as it is the first term that is expliicty said.
There are 4 terms overall.
Plugging it into the formula...
Sₙ = 4(1-4⁴)/1-4 = 4(-255)/-3 = -1020/-3 = 340
Why do we use this ? This is just an easy example, some series can be absolutely crazy – this is what the series are for.
Hopefully that helps ! I only specified what the formula is and how it's used, not the background of it.(14 votes)
- Isn't the formula the same as a(r^n-1)/(r-1)? Isn't that just simpler?(7 votes)
- At4:50,I dont understand why he added the two equations together.(5 votes)
- For the geometric series: a+ar+ar^2+...
the sum of the first two terms is 24 and the sum to infinity is 27. Show that r=+-1/3. Could anyone help with this, please?(4 votes)
- Nice problem!
The sum of the first two terms is a+ar = a(1+r) = 24.
The infinite sum is a/(1-r) = 27.
A good trick is to divide these equations, so that the factors of a cancel out.
Divide the first equation by the second equation to get (1+r)(1-r) = 24/27 which gives 1-r^2 = 8/9 which gives r^2 = 1/9 which gives r = +-1/3.(3 votes)
- In another video Sal uses "r" instead of "-r" to multiply the Sn by and comes up with formula for a finite sum as a-ar^(n+1)/1-r instead of a-ar^n/1-r.
How can both be correct? Basicaaly, one formula has n and one has n+1.(4 votes)
- At2:42, why did Sal multiply the sum of all the terms by -r and not positive r? And what is the reasoning behind multiplying the sum of all the terms by r in the first place?(4 votes)
- When he multiplied by -r and then "added" them together it made Sn + -rSn have a bunch of terms cancel out since a bunch became negative, so this way you are left with a much more simplified.
Let me know if this doesn't answer your question and I can try something else.(1 vote)
- Shouldn't this video be put before the second video in this module?(5 votes)
- I would say it is the last one, because it's an extra just for those curious about math in general. You aren't really obliged to know the logic behind this formula as it is shown in this video (even though it is what makes math beautiful) as long as you memorize it and apply it correctly.
That's just my justification of this particular order of videos in this module.(0 votes)
- Let me get this straight. This first term is found when the index is 0? It's confusing when you're thinking about the first term and the 0, because I want to think the zeroth term. Is this correct or am I wrong?(2 votes)
- The first term is the first term. It has nothing to do with what the index is. There is no such thing as a "zeroth term".(2 votes)
- At4:25, Sal multiplies ar^(n-1) by -r and gets -ar^n. I do not quite get how that works and would like some help on it.
Thanks in advance!(2 votes)
- OK, this is a really REALLY great question. When you multiply ar^(n-1) and -r together the first thing you can do is distribute the negative sign, which gives you -ar^(n-1) * r. The variable r can also be expressed as r^1. So you get -ar^(n-1) * r^1. Next you can pull out the -a which gives you (-a)(r^(n-1)) * r^1. Then you can simplify and get (-a)(r^(n-1+1)). Once again that can be simplified very easily to
-ar^n. I hope that was helpful.(2 votes)
- Let's say we are dealing with a geometric series. There are some things that we know about this geometric series. For example, we know that the first term of our geometric series is a. That is a first term. We also know the common ratio of our geometric series. We're gonna call that r. This is the common ratio. We also know that it's a finite geometric series. It has a finite number of terms. Let's say that n is equal to the number of terms. We're going to use a notation S sub n to denote the sum of first. n terms. The goal of this whole video is using this information, coming up with a general formula for the sum of the first n terms. A formula for evaluating a geometric series. Let's write out S sub n. Just get a feeling for what it would look like. S sub n is going to be equal to, you'll have your first term here, which is an a and then what's our second term going to be? This is a geometric series so it's going to be a times the common ratio. It's going to be the first term times the common ratio. The first term times r. Now, what's the third term going to be? Well, it's going to be the second term times our common ratio again. It's going to be ar times r or ar squared. We could go all the way to our nth term. We're gonna go all the way to the nth term and you might be tempted to say it's going to be a times r to the nth power but we have to be careful here. Because notice, our first term is really ar to the zeroth power, second term is ar to the firsth power, third term is ar to the second power. So whatever term we're on the exponent is that term number minus one. If we're on the nth term it's going to be ar to the n minus oneth power. We want to come up with a nice clean formula for evaluating this and we're gonna use a little trick to do it. To do it we're gonna think about what r times the sum is. We're gonna subtract that out. We're gonna take the r times that sum, r times the sum of the first nth terms. Actually, let's just multiply negative r. Something that we can just add these two things and you'll see that it cleans this thing up nicely. So what is this going to be equal to? This is going to be equal to, well if you multiply a times negative r, we will get negative ar. I'm just gonna write it right underneath this one. So if you multiply this times negative r. I'm just gonna multiply everyone of these terms by negative r. That's the equivalent of multiplying negative times the sum. I'm distributing the negative r. If I multiply it times this term, a times negative r, that's going to be negative ar. Then, if I multiply ar times negative r that's going to be negative ar squared. You might see where this is going. And just to be clear what's going on, that's that term times negative r. This is that term times negative r. And we would keep going all the way to the term before this times negative r. So the term before this times negative r is going to be, let me put subtraction signs, it's going to be negative a times r to the n minus one power. That was the term right before this. That was a times r to the n minus two times negative r is gonna give us this. It's gonna get us right over there and then finally we take this last term and you multiply it by negative r, what do you get? You get, negative a times r to the n. You multiply this times the negative, you get the negative a and then r to the n minus one times r, or times r to the first, well this is going to be r to the n. Now what's interesting here is we can add up the left side and we can add up the right hand sides. Let's do that. On the left hand side we get, S sub n minus r times S sub n and on the right hand side we have something very cool happening. Notice, this a, we still have that. The a sits there but everything else, except for this last thing, is going to cancel out. These two are gonna cancel out. These two are gonna cancel out. All we're gonna have left with is negative ar to the n. It's going to be a minus a times r to the nth power. Now we can just solve for S sub n and we have our formula, what we were looking for. Let's see, we can factor out an S sub n on the left hand side. You get an S sub n, the sub of our first n terms. You factor that out, it's gonna be times one minus r is going to be equal to and on the right hand side we can actually factor out an a. It's going to be a times one minus r to the n. To solve for S sub n, the sum of our first n terms, we deserve a little bit of a drum roll here, S sub n is going to be equal to this divided by one minus r. It's going to be a times one minus r to the n over one minus r. And we're done. We have figured out our formula for the sum or for the sum of a finite geometric series. In the next few videos or in future videos we will apply this and I encourage you, whenever you use this formula it's very important, now that you know where it came from, that you really keep close track of how many terms you are actually summing up. Sometimes you might have a sigma notation and it might start it's index at zero and then goes up to a number, in which case you're gonna have that number plus one term. So you're going to have to be very careful. This is the number of terms. This is the first term here, we define it up here. N is the number of terms, the first n terms, r is our common ratio.