Integrated math 3
A finite geometric series can be solved using the formula a(1-rⁿ)/(1-r). Sal demonstrates how to derive a formula for the sum of the first 'n' terms of such a series, emphasizing the importance of understanding the number of terms being summed.
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- 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!(63 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.(118 votes)
- Why is it that I was watching a video in which he says he already derived this formula, and when I finally found the video where he derives it, it's located after the video I was watching, in which it was assumed I knew this formula.(25 votes)
- This lessson should be placed higher up right after "Geometric series with sigma notation" because in the video lesson following "Worked example:finite geometric series(sigma notation) it says the general formula was already mentioned in a previous video when it was not.It is only mentioned in this last video lesson "Finite geometric series with formla justification".
Please correct this mistake as it is confusing.(26 votes)
- Great, so, that's the formula. Simple. But WHY? Why does this formula give us the sum? Does anyone know of any videos anywhere that actually explain WHY this works? And where it came from? Sal said "We're going to think about what r times the sum is and then subtract that out" but never gave an explanation.
ETA: If anyone's interested, I just found an awesome vid on Eddie Woo's youtube channel that goes into more detail on why this works, it's called: Intro to Geometric Progressions (2 of 3: Algebraic derivation of sum formula)(34 votes)
- This is just what I came here to post. It's as if we're supposed to just say, "cool thanks for the formula Sal!" and walk away without actually understanding what exactly is going on under the hood here... Thank you for the link I'm going to check it out now(6 votes)
- this is where I still struggle... how do I know to multiply by -r and then add the resulting equation to the original? I guess it's just, well that, a guess and it's "intuition", but then... my question is how do I get to build that intuition so that I can do it myself for other things?(23 votes)
- Practice helps build intuition, now for an endless amount of series to practice with I can only highly recommend pascal's triangle, and using its "diagonals" as series and trying to figure out the formula for each of them.
Here's a picture of pascal's triangle, and the "diagonals" are highlighted http://www.mathsisfun.com/images/pascals-triangle-2.gif
to make pascal's triangle you start with 1
For each consecutive row you add the number on the left and the right on the rows above to get your number, and a blank = 0... I can't explain it properly but its super easy, so here how it goes :
Row 1 = 1
row 2 = 0+1 , 1+0 = 1 , 1
row 3 = 0+1 , 1+1 , 1+0 = 1 , 2 , 1
row 4 = 0+1, 1+2, 2+1, 1+0 = 1 , 3 , 3 , 1
The diagonals are:
D1 = 1, 1, 1, 1, 1, 1, 1, 1 ...
D2= 1, 2, 3, 4, 5, 6...
D3= 1, 3, 6, 10, 15 ...
Try taking the sum of these series, and make a function for each of them, and then find a generic formula for all the diagonals if you're feeling brave!
A tip i can give you, is to try to go from something you don't know to something you do know, the path between the two is "intuition".
And as a bonus, pascal's triangle has way more than just series, try exploring it and figuring out its properties, it's fascinating ! By doing so, you'll be building up your "intuition", I can guarantee it! if the greeks had known about it, they'd have built temples and revered it like a deity.(19 votes)
- Is it possible to find n by using a formula, as it is with arithmetic series?(4 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(0 votes)
- Is there a name for this technique of finding a formula?(11 votes)
- Converting between a recursive form and an explicit form of an expression?(0 votes)
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)
- How am I supposed to memorize this formula? I understand everything in the video but it just doesn't stick to my head. I was reviewing this lesson a few weeks after I learned it and I didn't remember anything. Also, when am I supposed to use this formula? I don't really understand its purpose.(3 votes)
- You don't need to memorize it. Practice questions involving this formula and you'll eventually remember it. Plus, the derivation isn't too hard, so even if you forgot the formula, you can just derive it.
The formula is used a lot in infinite series, where we have infinite geometric series which converge. There, you use a slightly modified version of this formula to find the sum. This formula itself is used for, as the video says, finding the sum of a finite geometric series(5 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.