Log in averynash 8 years agoPosted 8 years ago. Direct link to averynash's post “At 4:25, Sal multiplies a...” At Thanks in advance! • (67 votes) yeloc1 8 years agoPosted 8 years ago. Direct link to yeloc1's post “OK, this is a really REAL...” 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 (135 votes) Eliza 4 years agoPosted 4 years ago. Direct link to Eliza's post “Great, so, that's the for...” 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) • (45 votes) mareli vaneti 4 years agoPosted 4 years ago. Direct link to mareli vaneti's post “This is just what I came ...” 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 (8 votes) mason.acevedo2000 7 years agoPosted 7 years ago. Direct link to mason.acevedo2000's post “Why is it that I was watc...” 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. • (26 votes) Behrooz Rahn 7 years agoPosted 7 years ago. Direct link to Behrooz Rahn's post “This lessson should be pl...” 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". (29 votes) Jorge R. Martinez Perez-Tejada 8 years agoPosted 8 years ago. Direct link to Jorge R. Martinez Perez-Tejada's post “this is where I still str...” 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? • (24 votes) Yoann Nouveau 8 years agoPosted 8 years ago. Direct link to Yoann Nouveau's post “Practice helps build intu...” 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 The diagonals are: 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. (20 votes) Sevil Hikmatova 6 years agoPosted 6 years ago. Direct link to Sevil Hikmatova's post “Is it possible to find n ...” Is it possible to find n by using a formula, as it is with arithmetic series? • (4 votes) cossine 2 years agoPosted 2 years ago. Direct link to cossine's post “The video is actually abo...” 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) john 7 years agoPosted 7 years ago. Direct link to john's post “Is there a name for this ...” Is there a name for this technique of finding a formula? • (11 votes) Home Instruction and JuanTutors.com 6 years agoPosted 6 years ago. Direct link to Home Instruction and JuanTutors.com's post “Converting between a recu...” Converting between a recursive form and an explicit form of an expression? (1 vote) 𝘽𝘼𝙏𝙈𝘼𝙉 4 years agoPosted 4 years ago. Direct link to 𝘽𝘼𝙏𝙈𝘼𝙉's post “*QUESTION*ok... I'm so ...” QUESTION 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) A/V 4 years agoPosted 4 years ago. Direct link to A/V's post “So the majority of that v...” 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 As an example: 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) amydylee a year agoPosted a year ago. Direct link to amydylee's post “How am I supposed to memo...” 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. • (5 votes) Venkata a year agoPosted a year ago. Direct link to Venkata's post “You don't need to memoriz...” 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 (6 votes) Mikey Schwartz 7 years agoPosted 7 years ago. Direct link to Mikey Schwartz's post “Isn't the formula the sam...” Isn't the formula the same as a(r^n-1)/(r-1)? Isn't that just simpler? • (7 votes) Kim Seidel 7 years agoPosted 7 years ago. Direct link to Kim Seidel's post “Yes, the 2 formulas are t...” Yes, the 2 formulas are the same. It is your choice as to which you think is simpler. (3 votes) Stephen 4 years agoPosted 4 years ago. Direct link to Stephen's post “At 4:50,I dont understand...” At • (6 votes) Ian Pulizzotto 4 years agoPosted 4 years ago. Direct link to Ian Pulizzotto's post “He added the equations so...” He added the equations so that most of the terms on the right hand side would cancel out. (4 votes)Want to join the conversation?
4:25
-ar^n. I hope that was helpful.
Please correct this mistake as it is confusing.
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
etc
D1 = 1, 1, 1, 1, 1, 1, 1, 1 ...
D2= 1, 2, 3, 4, 5, 6...
D3= 1, 3, 6, 10, 15 ...
etc
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!
r = the common ratio
n (exponent) = number of terms.
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. 4:50
Video transcript
- 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 ratioof our geometric series. We're gonna call that r. This is the common ratio. We also know that it's afinite geometric series. It has a finite number of terms. Let's say that n is equalto 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 videois using this information, coming up with a generalformula for the sum of the first n terms. A formula for evaluatinga geometric series. Let's write out S sub n. Just get a feeling forwhat 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 firstterm times the common ratio. The first term times r. Now, what's the third term going to be? Well, it's going to bethe 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 tobe 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 witha nice clean formula for evaluating this and we're gonna use a little trick to do it. To do it we're gonna thinkabout 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 canjust add these two things and you'll see that itcleans 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 itright 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 thisterm, 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 timesnegative 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 nminus 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 canadd 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, thesub 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, thesum of our first n terms, we deserve a little bitof 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 outour 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 thisformula it's very important, now that you know where it came from, that you really keepclose 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 haveto be very careful. This is the number of terms. This is the first term here, we define it up here. N is the number ofterms, the first n terms, r is our common ratio.
