Algebra (all content)
- Arithmetic series intro
- Arithmetic series
- Worked example: arithmetic series (sigma notation)
- Worked example: arithmetic series (sum expression)
- Worked example: arithmetic series (recursive formula)
- Arithmetic series worksheet
- Arithmetic series
- Proof of finite arithmetic series formula
Sal evaluates the arithmetic series Σ(2k+50) for k=1 to 550. He does that using the arithmetic series formula (a₁+aₙ)*n/2.
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- Is that formula the same for all sigma notation problems?(7 votes)
- I'm assuming you are referring to the formula for the sum of a finite arithmetic series, which Sal defines starting at around2:10. If that's the formula you mean, then: No, it isn't the same for all finite sums: The formula Sal uses will work only for arithmetic series. Sigma notation is used for all kinds of sums, and not just arithmetic series. Different series will have different sum formulas.(9 votes)
- Why does Sal add two sums of S(n)?(4 votes)
- I believe he was trying to show us why we have (n/2) and how the final answer comes about. He explains where the /2 comes from in the Arithmetic Series Formula video so if the formula confuses you best to probably rewatch that video, but as you can he gives you the answer at4:30so everything else after that is just him proving it to us.(4 votes)
- I checked that it is an arithmetic sequence first by seeing that d( 2*x + 50 ) / dx = 2 which is a constant. Thus the sequence terms are changing by a constant. Does this mean that we get an arithmetic series whenever the expression is linear with respect to the index ?(3 votes)
- Yes, because the "𝑥:th" term of an arithmetic sum is always 𝑡(𝑥) = 𝑎 + 𝑏𝑥, where 𝑎 = 𝑡(1) and 𝑏 is the difference between two consecutive terms, 𝑏 = 𝑡(𝑘 + 1) – 𝑡(𝑘).
This means that the sigma notation will be 𝛴(𝑎 + 𝑏𝑥), 𝑥 = 0 → 𝑛 – 1, where 𝑛 is the total number of terms.
Even if we changed the notation to, for example, 𝛴(𝑎 + 𝑏(𝑥 – 1)), 𝑥 = 1 → 𝑛 ⇒
⇒ 𝑑/𝑑𝑥 ∙ (𝑎 + 𝑏(𝑥 – 1)) = 𝑏, so it still holds up.
And, since these are all definitions of an arithmetic sum, there can't be any other types of sums that match this relationship.(5 votes)
- Well, I what I don't get is that in the video 'Worked example: arithmetic series (recursive formula)' the rule is that it increases with a + 11. And a = 4. But in that video Sal said the 1st term would be 4, not 15 which is 4+11. How come here he actually works out the 1st term using the rule so the 1st term is 52 which K = 1 *2 + 50? Why can't he just be consistent? :((3 votes)
- It's not a matter of being consistent. He's right. The first term in Worked example: arithmetic series (recursive formula) is 4. Then the second term was 15. This is because it's a recursive. He was given that the first term is 4. Now for this video, it's the sum of 2k + 50 from index k =1 to k=550. Since the index starts at k =1, he has to plug in 1 to find the first term so, it's 52. If you were thinking that you plug in 0 to find the first term is 50, you can't do that because the index starts at k =1. Hope this helps. If you need further clarification let me know.(3 votes)
- Does n in the formula for S_n strictly represent the number of terms being summed, or the value of the upper index? I'm asking because I've noticed that the starting index doesn't have to be 1.(2 votes)
- In the event of S(n), "n" is going to refer to the number in a series, and nth term of it. In the case of sums of series, you will commonly see sigma notation used, where the starting term does not have to be the first term in a series. In that case, expect to find the the number of the term that "starts" the addition below the sigma, and the final term on top.(4 votes)
- In this example, "k" is 1-550, which means "n" of the sum formula would be 550 terms, correct? If "k" was 0-550, would "n" be 551 terms?(2 votes)
- k varies, starting at 1 and going up to 550. How many values of k is that?
If j varies from 1 to 5, how many values does it take?
So therefore, in the video, k takes 550 values.
If j goes from 0 to 5, how many values does it take on?
(0, 1, 2, 3, 4, 5; that is, 6 values.)
So if k goes from 0 to 550, yes, that's 551 values.
B u t, don't forget that n is still 550, not 551.(3 votes)
- Isn't this formula that the legendary mathematician Gauss used in elementary school to shock his teacher?(3 votes)
- Yeah, the story goes that the teacher told the students to calculate 1 + 2 + 3 + ... + 100, which the teacher thought would take a while, but Gauss (10 years old at the time) quickly realized that it added up to 101 ∙ 50 = 5050, which, needless to say, startled the teacher.(2 votes)
- Hey guys, I need some help! How do I simplify this?
An auditorium can fit 252 people. If the first row fits 16 people and the following rows can fit 3 more people than the preceding row, how many rows are there in the auditorium?
Sn = n(a1+an)/2
252 = n(16 + an)/2
Because I don't know a sub n, I can't find n, so I have to plug in numbers till I find the answer:
Thus, there are 9 rows. Is there an equation I can use to solve for this without plugging in numbers?(2 votes)
- One method is to get an in terms of n. Since the first term is 16, and each term after the first is 3 more than the previous term, it follows that the nth term is an = 16 + 3(n - 1).
So we have the quadratic equation 252 = n[16 + 16 + 3(n - 1)]/2, which we can then solve for n. While the quadratic could be solved by factoring, using the quadratic formula is probably easier in this situation. Note that we would disregard any solution that is negative or fractional, because the number of rows, n, must be a non-negative integer.
Have a blessed, wonderful day!(3 votes)
- Why do we divide by 2?(2 votes)
- [Voiceover] So I have a finite series here expressed in sigma notation and I encourage you to pause the video and see if you can figure out what this evaluates to. This is going to evaluate to a number. So assuming you've had a go at it, let's work through this together. So this is a sum from k equals 1 to k equals 550, so we're going to have 550 terms here and it's the sum from k equals 1 to k equals 550 of 2k plus 50, so whenever I try to evaluate a series, I like to just expand out the sum a little bit just so that I can get a feel for what it looks like, so let's see. This is going to look like, when k is equal 1, this is going to be 2 times 1 plus 50. When k is equal to 2, it's going to be 2 times 2 plus 50. When k is equal to 3, it's going to be 2 times 3 plus 50. And we're going to keep going all the way until we get to the last term, when k is equal to 550, it's going to be 2 times 550 plus 50. So let's see. This first term is going to be, it evalutes to 52 plus, this next term is 2 times 2 plus 50 is going to be 54, plus the next term, 2 times 3 is 6 plus 50 is 56, and we're going to go all the way, all the way to our last term, 2 times 550 is 1100 plus 50 is going to be 1150. So that gives us a good feel for this sum, for this series. We're going to start 52, and we're just going to keep adding 2 for each successive term, all the way until we get to 1150, and we're going to take the sum of all of these, and since each successive term, we're increasing by the same amount, we're increasing by 2, we're increasing by 2, we can recognize this as an arithmetic series. We are increasing by the same amount each time. And there is a formula for the sum of an arithmetic series, and first we're just going to apply the formula, but then we're going to get a little bit of an intuitive sense for why that formula works, and actually, in other videos, we have proved this formula, but it's always good to get a sense that, you know, that this formula just doesn't come out of thin air, so the formula for the sum of an arithmetic series, so the sum of the first n terms is going to be the first term plus the nth term over 2, so it's really the, it's really the arithmetic mean of the first and the last terms, you could say the average, in, I guess, everyday language, average of the first and last terms and then times the number of terms you actually have, so if we were to try to apply it to this case, we're trying to take the sum, sum of the first 550 terms, I'll do this in a new color just for kicks. All right, so we're going to take the sum of the first 550 terms, and it's going to be equal to the first term, so that's 52 plus the last term, the nth term, 1150, it's really just the average of those two, the average of the first and the last term and then times the number of terms we have, times 550, so what is this going to be? Well, we could simplify this a little bit. If we're going to take 550 divided by 2, this is going to be, I could write this as times, actually, let me just, let me just simply this in a different way. So this is going to be the same thing as, I could write this, 52 divided, well, let me just add first. This is going to be 1,202 over 2. All right, did I do that right? Yeah, 1,202 over 2 times 550. Now, 1202 divided by 2 is going to be 601, so this is equal to 601 times 550. And let's see, I can multiply that out. So, let me just do 550 times 601, so 1 times 550 is 550, and then I have a zero here, but I just have a zero there, so zero times 550, I'm just going to get a bunch of zeros, and then I go to the hundreds place. 6 times zero is zero. 6 times 5 is 30. 6 times 5 is 30 plus 3 is 33. You add it all together, we get a zero, we get a 5, we get a 5, we get a zero, we get a 3, we get a 3. We get 330,550. That's what this whole thing, that's what this whole thing sums up to. Now, I just said that we'll get a little bit of intuition for why we were able to just apply this formula, and let's just think about what the sum of the first 550 terms is, and I just wrote it down up here. So, let me write it, I'm just going to switch colors again. So, we're going to have the sum of the first 550 terms, which is what we just wrote over here, we already said this is going to be 52 plus 54 plus 56 plus, and we're going to keep going all the way to 1150. I'm going to write it again, the sum of the first 550 terms, but I'm going to just write it in reverse. We can obviously swap the order in which we add. It's going to be 1150 plus 1150 minus 2, which is 1148, plus that minus 2, which is 1146, and we go all the way to the first term, all the way to 52. Now, what I want to do is just I want to add these two sums, so I'm just going to get 2 times the sum of the first 550, so if I had the two left sides, I would have 2 times the sum of the first 550 terms, and we do this generally when we prove this formula in previous videos, but I always like to get a sense of where it comes from and so this is going to be equal to, well if I add these two terms right over here, I get what? I get 1202. That number should look familiar. And then if I add these two, right over here, what do I get? I get 1202. And then if I add, I think you see where this is going, and then if I add these two characters, what do I get? I get 1202, all the way to these last two characters. You add them together, what do you get? You get 1202, so how many 1202s do I have? Well I have 550 of them. There are 550 of these terms. So this is going to be equal to 550 times 1202. And so if you just wanted to solve for this sum, you just divide all these sides divided by 2, so you divide by 2. You divide by 2. You divide by 2, and that's exactly what we did over here. 550 times 1202, divided by 2. So hopefully that gives you an intuition for things.