 Hi, I'm Zor. Welcome to Unisor Education. I would like to offer you a few problems about series. This is the second set of problems in the same topic and sequences and series. Usually the problems in this particular topic are very, very simple. Like, okay, here is the geometric sequence. Here is the first element of this sequence. This is the quotient. Please calculate using the formula, the sum of the first 100 elements or something like this. Well, that's not an interesting kind of problems. I wanted to present to you something which is a little bit more challenging. So please bear with me. They are not difficult but unusual, I would say. And yeah, that might be actually challenging for some of you. Which is the purpose of this course anyway, to challenge your mind, right? Alright, starting from the beginning. First is a relatively easy problem which I have already mentioned in one of my theoretical lectures. If you sum the first n odd numbers, you will get n squared. Alright, so 1 plus 3 plus 5 plus etc plus... My nth odd number is 2n minus 1. So if you summarize, it would be n squared. Well, indeed, the first one will be 1, which is 1 squared. For the first two, it will be 1 plus 3 is 4, which is 2 squared. For the first three, it will be 1 plus 3, which is 4 plus 5 is 9, which is 3 squared, etc. So we have to prove this formula. We can prove it in two different ways. One is, I would say, straightforward. You can just prove it by induction. Now, obviously, it's true for n equals 1. Now, we assume that for n equals k, it's true. So when a is equal to k plus 1, we have one extra element, which is the sum of the first k element, which we assume is k squared. Plus the next element, which is 2k plus 1, right? If the case element is 2k minus 1, then the next odd number would be 2k plus 1. And obviously, this is k plus 1 squared, which basically proves our formula. So for n equals 2k plus 1, the result is k plus 1 squared. Slightly different, not maybe as straightforward solution is to use the same methodology. I was using to derive the formula for arithmetic progression. So let's just have the same sum in reverse, starting from 2n minus 1. Then it would be 2n minus 3, the previous odd number, etc. And the last one would be 1. So that's exactly the same as if we sum them up together. We will have the first one and the last one would be 2n, obviously. Then the second and the second from that side would be again 2n, etc., etc. So every one of them 2n minus 1 plus 1 would be 2n. The number of these would be obviously n, n components in this sum, which means that 2s is equal to n times 2n, which is 2n squared. So the s is equal to n squared. That's another way to prove it. And finally, obviously, you can use the formula for sum of arithmetic progression with the first element 1 and the difference 2 in this particular case. Substitute all these guys into the formula and you will also get n squared, obviously. Now, but I never remember the formula for sum of arithmetic progression or geometric for the same token. I do remember the way how I derive the formula by summing up in a reverse order. Basically, if you remember this methodology, to come up with a formula is really like a one-minute deal. So, different people approach it different. Some people prefer to memorize the formula. I don't memorize the formulas. I memorize the way how these formulas are derived. So that's what actually I... Well, I'm not really saying that you should do it this way. You do whatever you want. But, well, it's probably very individual. Some people prefer to memorize the formula. Some people prefer to memorize the way how it was derived. And in this particular case, by the way, the methodology of derivation of the formula by just summing up in reverse is very simple to remember. The formula is, well, for me, it's much more complex. So, basically, we have proven that the sum of n odd natural numbers gives you n squared, n of problem. Next. Find the elements of arithmetic progression if there's sum... Find four elements of arithmetic progression. If there's sum is one and sum of their cubes is 0.1. All right. So we have four members, four elements of arithmetic progression, which is a. Next is a plus d. Next one is a plus 2d. And next one is a plus 3d. Okay, four members of arithmetic progression, which basically have two variables which they depend upon. The first variable is a, which is the first element of this progression. The second element is d, which is the difference of arithmetic progression. Right? So these are four elements. Now, we know two things about these four elements. That their sum is equal to one and the sum of their cubes is equal to 0.1. So we have two variables and we have two conditions, which means we have actually a system of two equations with two unknown variables. So this is one equation and this is another equation. Well, quite frankly, it looks a little scary. This is the third power. You don't really know how to solve equations of the third power. However, these are certain specific conditions which will allow you to simplify the whole thing so you don't really have to solve the problems, the equations of the third degree. Now, what can we do about this? Well, first of all, I would probably find out what is a in terms of d from the first equation. Now, we have one, two, three, four. We have four a's, so it's four a plus d and two d and three d. That's three six. That's six d equals two one. From which I can derive that a is equal one minus six d over four. Now, I can substitute this a into these guys, right? Having only one equation with only one variable d. Now, let's do it. So I have one minus six d over four q plus a plus d. Let me just think about what it is. If I plus d here, I have four as a common denominator. It means it will be four d. So one minus six d plus four d, which is one minus two d over four q. Next, next is plus two d. Two d, four and two d is eight d on the top so it will be one minus six d plus eight d. So it will be one plus two d over four q. And the last one will be a plus three d. So if I have plus three d here, it's 12 d on the top in the numerator minus six d plus 12 d, so it's plus six d. So it's one plus six d over four q. And what is equal to zero point one? Now, I don't need this. The only thing which I do need is the fact that a is equal to one minus six d over four from the first equation. And this is how the second equation is transformed into when we are using only one variable substituting this value a into each member. Now, you see one minus six d, one plus six d. One minus two d, one plus two d. I will use a very easy formula. A cubed plus b cubed is equal to a plus b. a squared minus a d plus b squared. Well, for those who know about this, that's obvious for those who don't just multiply them and you will see that this is exactly what it is. Because this is a cubed and then you have minus a squared b and plus a b squared. This would be plus b, a squared b, and minus a b squared. So it will cancel out and only a cubed and d cubed will remain. So I will use this formula grouping these two guys and these two guys. Now, what do I have? So this plus this is a plus b. As you see, it will be one minus six d over four and one plus six d over four. Now, if I sum them together, my six d would cancel out because it's minus and plus and all I will have is a plus b. This plus this would be one fourth, right? Now, multiply by... Now, here I have to really do more or less one minus six d over four squared minus a b. I have to multiply this by this. It will be minus one minus six d, one plus six d divided by six d, right? Four and four. And finally, plus d squared, which is one plus six d squared divided by 16. I'll do it as well. Okay, that's my these two elements. Now I have to add... Now, one quarter is... Now I have to do the same with these two. So it's plus, again, the sum of these two a plus b, the sum, two d will cancel out, so I will have only one quarter plus one quarter. Oh, wait a minute. Did I make a mistake? Yes, it's two fourths, actually. One minus six d and one plus six d. Yes, six d will cancel out, but one plus one would be two, which means it's two fourths or one half. My mistake, sorry about this. Similarly here. Very similar, but instead of six d, I will put two d. Two d squared divided by 16 minus one minus two d, one plus two d, 16 plus one plus two d, square 16. The whole thing is equal to one tenth, right? So that's how my equation looks now. Is it simpler? Well, we have to basically simplify it a little bit more than that. Well, first of all, one half and this 16, one half and this 16, so it's one thirty second, actually, right? If I will take the sixteenths out of this, it will be one thirty second. So one tenth, this is one tenth, is equal to one thirty seconds off. And now I have these guys and these guys. Let me just open the parenthesis. One minus twelve d plus thirty six d squared, right? This is one minus six d squared. Plus, sorry, minus, minus. Now one minus six d and one plus six d is one minus thirty six d squared, so it will be minus one plus thirty six d squared. Now plus this, plus one plus twelve d plus thirty six d squared. Here, one minus four d plus four d squared minus one plus four d squared and plus one plus four d plus four d squared. Close parenthesis. This is one thirty second of the whole expression and that's what we have now. It's easier. Yes, it is. Because now lots of things will cancel out. Now what cancels out? Twelve d, twelve d, four d and four d. Plus one minus one minus one plus one. So what's left? One tenth is equal to one thirty second off. I have one and another one. So I have two. Now I don't have anything d in the first degree. I only have d to the second degree. So what I have, thirty six plus thirty six plus thirty six. It's three times thirty six which is one or eight and four and four and four which is twelve and it's one twenty. So one twenty d squared. That's what I have. Now do we see how easy it is? Well let's now simplify it a little bit. So we multiply this by thirty two. So we have thirty two is equal to ten times one two plus one twenty d squared. So it's two hundred plus twelve hundred d squared. Am I right? Yeah I think we can reduce it by two. So it's sixteen, actually by four. So it's eight. Now it's better to do it by two. So it's sixteen is equal to one hundred plus now I think I screwed up something. Now that's wrong. So it's two plus one hundred and twenty d squared is equal to thirty two over ten. Now if you look two which is so one hundred and twenty d squared is equal to thirty two over ten minus two which is twelve tenths, right? This is twenty over ten. Thirty two over ten minus twenty over ten is twelve over ten. And now, okay, now we can reduce it by a hundred and twenty to get the d squared. So if one hundred and twenty d squared is equal to twelve over ten d squared is equal to twelve over one hundred and twenty times ten which is reduced by twelve which is one hundredth. From which d is equal one tenths. Well isn't that wonderful? Such complicated things gives us one simple solution, one tenths. Now actually there is another solution to this. d equals minus one tenth. Well we will check if our checking procedure will produce something similar. Now let's talk about a now. So a is equal to one minus six times one tenths divided by four which is six tenths which is four tenths, four tenths divided by four which is one tenths. So for d equal one tenths, a equals one tenths. And I'm talking about a arithmetic progression which is zero point one, zero point two, zero point three and zero point four, right? The first element is zero point one and the difference is zero point one. So these are our four numbers. Let's check it out. Their sum is equal to zero point one plus two that's three tenths, that's six tenths, that's ten tenths which is one. That's exactly what we need. Now their cubes are zero point zero zero one, zero point zero zero eight, zero point zero twenty seven and zero point sixty four, right? Sixty four thousandths. This is nine, this is thirty six and this is one hundred one thousandths which is one tenths. So the sum of these is equal to one and sum of these is equal to one tenths. So the exactly whatever we need to do this is the solution. How about minus one tenths? Well let's just check it out. G is equal to minus one tenths. A is equal to one minus which is plus zero point six over four which is zero point six over four which is zero point four. So we have zero point four, zero point three, zero point two and zero point one which is basically exactly the same kind of a arithmetic sequence but in reverse. From the maximum and going with a negative difference to the beginning to the zero point one. And obviously since we got the same numbers we have all these equations satisfied. Their sum is equal to one and the sum of their cubes is equal to zero point one. So this is it. This is one of the problems which are not as trivial as typical problems about geometric and arithmetic sequences. And I have a couple of more. So let me just try one more thing and probably that would be it for this particular lecture. So what's my next? Finds of three sides of the triangle if they are integer numbers forming arithmetic progression and the perimeter of the triangle is 15 and find all the solutions. So I have a triangle. I know that all these lengths are integer numbers. The perimeter is 15. And I also know that they are constituting three numbers which are really an arithmetic progression which means starting from some number x and a difference d. I would have another x plus d and another x plus 2d. So what do I have? I have the perimeter which is x plus x plus d plus x plus 2d equals 15. Now this is what? 3x plus 3d is equal to 15 or 3x plus d equals 15. Now how can I resolve this? How can I find the solution if there are two unknowns in one equation? Well, very simple. I know that these are integer numbers which means x plus d is integer. So how many, obviously from here, how many integer numbers x and d, positive integer numbers obviously because we're talking about triangle and its lengths. So how many natural numbers x and d exist so that their sum is equal to 5? Well, it's actually very limited number. Let me just enumerate all of them and we will get all the solutions. So if x plus d is equal to 5 then either I have 1 comma 4 or 2 comma 3 or 3 comma 2 or 1 comma 4. So we have four different solutions. Zero cannot be here and negative numbers cannot be here. So I have four different pairs of x and d which means there are four different triangles. One with sides 1 plus 4, 5 plus 4, 9 and obviously the perimeter is 15. In this case, sides are 2 plus 3, 5 plus 3, 8. And 8 also give me 15 as a perimeter. Now 3 and 2 are 3 plus 2, 5 plus 2, 7. This is also a solution. And finally, sorry, I had to put 4, 1 instead of 1, 4. And finally I have 4 plus 1, 5 plus 1, 6. So these are different triangles which can be constructed using integer, positive integer numbers which constitute arithmetic progression and the perimeter is 15. That's easy. And a similar problem I think I have find the lengths of three sides of a triangle with an integer number forming a geometric progression and the product of their lengths is 216. So it's geometric progression which means if one is equal to a another is aq and the third one is aq squared and their product which is a times aq times aq squared is equal to 216. Well this is obviously aq times q cubed which is actually aq cubed and this is equal to 216. Now 216 is 6 cubed. 6 times 6 is 36 times 6 is 216. So aq is equal to 6. Now, same logic. How many different a and q exist? Natural numbers, positive integer which have a product of their multiplication 6. Well, let's think about it. 1 times 6 2 times 3 3 times 2 and 6 times 1, right? So 1 and 6 which means 1, 6 and 36. So these are lengths of the triangle. Is it possible to have these lengths of triangle? Oh, by the way, that's actually something which I had to check for the arithmetic progression as well. These cannot be the lengths of the triangle because as you know some of any two sides should be greater than the third one. Now, let me return back to the previous case. When I had 5 as a sum, right? So it's 1, 4 2, 3 3, 2 and 4, 1. Yeah. So in one case it would be 1 plus 4, 5 plus 4, 9, right? Well, this is not possible to have as a triangle. So this solution is not good. So I kind of rushed to the conclusion basically forgetting that we are talking about geometric properties. Now, 2, 3 would be 2 plus 3, 5 plus 3, 8. Same, same. Some of these 2 is smaller than this. Not good. 3, 2 would be 3 plus 2, 5 plus 2, 7. This is greater than this. This is good. And 4, 1 which is 4, 5 and 6. This is also good. This is also good. So I have only two triangles. Now, in this particular case, what did I have noticed? So this is not the right triangle because there is no triangle with these sides. Although it satisfies our equation, but we cannot really build a triangle because there is one more condition that the sum of any 2 should be greater than the third one. How about 2 and 3? So the first one would be 2 multiplied by 3 would be 6 multiplied by 3 would be 18. Not good. The sum of these 2 is smaller than this. 3 and 2, it would be 3 times 2, 6 times 2, 12. Not good. 3 plus 6 is smaller than 12. And finally, 6 times 1, 6 times 1, 6. So it's an equilateral triangle. And this is the only solution in this particular case. So you have to be very careful and look at the problem how it's presented more carefully, which I was wrong not to do it with the arithmetic progression among the sides. Because in this case, it's only one solution because three solutions are not good. Okay, I think I will finish for today and I will present some other problems in the next lecture. So thanks very much for listening to me. Don't forget that every lecture dedicated to problems must first be examined through the notes on this side and all the problems you should really try to solve yourself. Listen to the lecture next and after I finish, do exactly the same thing just by yourself. That would be the great help for you. Thanks very much. Bye.