 Hi, I'm Zor. Welcome to Unisor Education. I will continue talking about series and sequences in this particular lecture. I will present some information pertain to series of progressions, arithmetic progression and geometric progression. Now, it's actually the most simple cases, actually, of some nation, arithmetic or geometric progression. But at the same time, they are actually occurring quite often, and that's why we start with, this is a simple and very often the occurring examples of series. So without further ado, let's just introduce these two things. So first of all, about arithmetic progression. Arithmetic progression is characterized by the first element, A, and the difference between two consecutive elements, which means that after the first element, A, next one would be A plus G, next one would be A plus 2G, etc. And what is my element number K? It's A plus K minus 1G. This is my generalized K's element of arithmetic progression with the first element A and the difference G. Well, indeed, if K is equal to 1, this is 0, so it's A. If K is equal to 2, 2 minus 1 is 1, so it's A plus G. If K equals to 3, 3 minus 1 is 2, so it's A plus 2G, etc. Now, let's sum them up together. Before doing that, I will remind you the case with famous German mathematician Gauss or Paulus. And when he was a child, I think I actually told you before this story, when he was a child, he was asked to, well, the whole class actually where he participated, was asked to basically sum up all the numbers from one to a hundred. And he was very quick with an answer. And what he did was, instead of doing this, he did this. He wrote exactly the same thing but backwards. Hundred, ninety-nine, ninety-eight, three, two, one. This is the same sum, right? Which means if I will add them together, it will be double the sum which we need. But considering that I can sum up these whole numbers in any sequence because addition is commutative and associative, so we will sum up these two sequences vertically. This and then this and then this and then this and this and this. And as you see, in every case, I have one hundred and one. Now, how many times I have one hundred and one? One, two, three, four, a hundred. So my total, which is double sum, is one hundred times one over one, which is this. But this is the double of the sum. So a single sum is five thousand and fifty. That's his answer which he came up with very quickly. I will do exactly the same thing for arithmetic progression in a more generalized case. Let's say I have to sum up all the numbers of arithmetic progression with numbers from one to some maximum number n. So my first number is a. My second number is a plus d. My third one is a plus 2d, et cetera. Now, my last number, number n, would be a plus n minus 1d. Now, the one preceding this would be 1d less than this. We're going backwards. So it's n minus 2d. So these are the beginning and the end of the sequence, which I have to summarize. This is s. So I'll put plus signs here. Now, I will do exactly the same thing as Gauss did for summation of first hundred numbers. I'll put them in reverse. So first, I will put this one, a plus n minus 1d. Secondly, I'll put this one, the previous one, a plus n minus 2d, et cetera. And the last one would be a. And the one before that would be a plus d. And now I have to sum them up together. And let's do it vertically again. The first one and the first one from the opposite side. The second one and the second one. Now, et cetera, et cetera. The last one with the first and the one for less with the second. And now I will, obviously, this is also s. So instead of summing sequentially, I will sum up vertically. And every pair, a plus this, what will be? Would be a and a would be 2a plus n minus 1d. Now, what will be here? a and a will be also 2a. n minus 2d and d would be again n minus 1d. Same thing. And as you guessed, everything else would be also a. I mean, 2a plus n minus 1d. And how many of these I have? Well, obviously, number n of the elements which I am summing up. So basically what I came up with here is that this sum can be calculated as n times a plus n minus 2a plus n minus 1d. And obviously, the formula for a sum s would be n 2a plus n minus 1d divided by 2. So this is a generalized formula for sum of the first n elements of arithmetic progression with the first member a and the difference d. Well, can it be simplified somehow? Yeah, I guess it can, but I don't think it matters. You can rewrite this as you can divide this by 2. So it will be a times n plus n n minus 1d divided by 2. Doesn't really matter. I mean, whatever way you write it, you will get exactly the same result. And this is, let me just use the summation in this case. My generalized element is a plus i minus 1d. So I'm summing up this with i is equal from 1 to n. So sum of all elements which look like this where i is an index changing from 1 to n is equal to this formula. Now, is it really a rigorous proof? Well, not exactly, because I was dealing with, well, basically some kind of a number. And I reversed the sequence. I think more rigorous proof would be just to do more or less the same thing by induction. And let me just, for an illustration, basically, do this again using the method of mathematical induction. So let me rewrite this formula again as this divided by 2. This is s n. I use the index n to signify that this is the sum of the first n elements of my sequence. So how the classical method of induction is supposed to be used? First, you check it for some initial value. Let's say n is equal to 1. Now, if n is equal to 1, my sequence contains only one element a. And obviously, the sum of this one element a is exactly the same thing a. So I should have s1 equal to a. Now, let's check it out. If n is equal to 1, this is 0, right? And this is 1. So I have 2a divided by 2, so it's a 5. So formula is correct for n equals to 1. Now, let's assume that for some n equals to k, s k is indeed equal to n times 2a plus n minus 1. I mean k. I'm using k here now. And this is also k. d divided by 2. Now, what happens if I will change to k plus 1? Now, what happens with my sum? Obviously, sum will increase by k plus first member of the sequence, right? So s k plus first is supposed to be equal to s k plus k plus first sequence. Now, k plus first sequence would be what? Remember, my generalized formula is this one, right? This is k's member. Now, if instead of k, it's k plus 1, it would be a plus k d. Now, this is how it's supposed to be, right? So let's check that if I will do this, I will have a similar formula to this one with n equals to k plus 1. All right, so how can we check that very simply? So I have to add k times 2 a plus k minus 1 d divided by 2. That's my s a plus a plus k d. Well, obviously, we will use the common denominator. So I will have, so this is s k plus first. So I will have k times 2 a plus k minus 1 d plus 2 a plus 2 k d divided by 2, right? OK, equals. Instead of k times 2 a plus 2 a, k times 2 a plus 2 a plus k times k minus 1 d plus 2 k d. OK, I have rewritten it in this way. I multiply k by 2 a separately than this member, because I will combine them together in the future. And k is multiplied by this and then this. So what do I have? OK, k times 2 a and 2 a would be, obviously, k plus 1 to a, right, plus. Here, if I will factor out k, I will get k minus 1 d plus 2 That's what I have. Equals k plus 1 to a plus k, k minus 1 times d and 2 d. It's k plus 1 d divided by 2. Equals. Now I can factor out k plus 1. And inside the parentheses, I will have plus k d divided by 2. Now, this is my final formula for s k plus k plus first member of this sequence. Now, if you substitute instead of n k plus 1, you will get exactly the same, you see? n minus 1 where k is equal to, when n is equal to k plus 1, would be exactly k. So my formula is exactly the same, but with n equals to k plus 1 instead of n equals k. That's the real proof. Now, what's wrong with this proof? Well, nothing wrong except that I kind of knew the formula, right? And now, knowing the formula, I can prove it by induction. But in the beginning, I didn't know the formula. I basically derived it using not exactly rigorous methodology of reversing the sequence and sounding it vertically. But whatever it is, I should probably say that the combination of both have intuitive proof, which is the first part, and the rigorous proof by induction, which is the second part. Together, they made the whole theory, well, relatively rigorous, all right? OK, this is all for arithmetic progression. Now, let me address the geometric progression, the sum of the first n members of geometric progression. Now, what is geometric progression? If you remember, it's characterized by the first member and the quotient, which is multiplied on every step. And the element number k would be this. So my question is, what is sk equal to a plus aq plus blah, blah, plus aq k minus 1? What is this sum? That's what I have to calculate. Well, I'll do exactly the same thing. I'll do first some kind of a trick which would help me to derive the formula. Now, what's the trick? The trick is very simple. I multiply by q this sum. So q times sk is equal to a times q is aq. a times q is aq squared. aq to the k minus 1 times q would be aq to the k's, k degree. And what is the previous member? Obviously, aq to the k minus 1. So my power is from 0 to k minus 1 in this case. But from this case, it's from 1 to k. Now, these two expressions are very much alive. Because as you see, this and this and everything in between are common with these guys, you see? So if I will do sk minus q times sk, what happens? If I subtract from this, I subtract this. Well, these guys will go, obviously. And what I will have is a minus aq to the k's. This is the only thing remaining from this. And this one with a minus sign because I'm subtracting is what remains from the second expression. And obviously, from here, I conclude that's minus minus q is equal to a minus aq to the k's, from which sk is equal to a, a minus q to the k's degree, 1 minus q. I factor out a. So I have 1 minus qk. And I divided both sides by 1 minus q. I get this one. So this is the formula. Now, if I really want to do it very, very rigorously, I have to, more or less, repeat my logic which I did with arithmetic progression, summing up arithmetic progression. I have to repeat here and prove it by induction. Well, I leave it as an exercise to you because it's extremely simple thing. So you can consider this as a final formula. What's also interesting here is another formula which you can actually derive from this one, a1 minus qk divided by 1 minus q. Now, the interesting formula which I wanted actually you to remember, it's more or less in line with, you remember the formula a minus b times a plus b is equal to a square minus b square. It's some kind of a very fast kind of a formula which you probably remember. Now, this is not as short, but it's also a very useful formula. Why don't you put a is equal to 1 in both cases? This formula is true for any a and any q, more or less. So let's assign a equals to 1. So what happens? Well, 1 minus q to the k is equal to 1 minus q, 1 plus q plus et cetera, plus q to the k minus 1. So if you have an expression like this, you can always represent it as the result of multiplication of these two expressions, which means that basically 1 minus 1 half to the fixed degree is equal to 1 minus 1 minus 1 half times 1 plus 1 half plus 1 half square plus 1 half q plus 1 half to the force degree. So this is an interesting formula which you can actually use for certain problems in a completely different area, like solving equations or anything like that. It allows you, basically, if you want to solve equation like this, you immediately see that q is equal to 1 is your first root of this equation. And then everything else is by one factor less. So for instance, you would like to solve an equation 1 minus x cubed is equal to 0. Well, I mean, yes, you can use some other methodology. But this is immediately, like x is equal to 1 is 1 root of this particular equation, 1 solution. And 1 plus x plus x square equals to 0 is another equation which you have to solve. And whatever the solutions of these plus x equals to 1 are the solutions of the whole thing. All right. Well, that's it for geometric progression. I will talk about certain problems related to arithmetic and geometric progressions in a separate lecture, including I will prove that, let's say, sum of arithmetic progression which contains odd numbers is a square of the last number. And I will talk about paradox offered by Greek philosopher Zina about Achilles and the tortoise. But so far, this is an introduction to what is exactly a arithmetic series and geometric series. I offered intuitive formulas, which I have derived with not exactly in a rigorous manner. But then you can prove both arithmetic series and geometric series formulas using mathematical induction, which I did for arithmetic case. And I do recommend you to use the same approach and prove by induction the formula for geometric series. That's it for today. 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