 Hello friends and welcome to another session of sequence and series and Continuing with our previous topic wherein we were trying to figure out a function which can help to Find out the nth term in a given sequence and we are trying to analyze a particular type of sequence only not that it's no You know, so it's not a random sequence. So there is a particular trend in this these the terms and In the previous two sessions, we saw how Sequence which can be given by a linear formula and then another sequence in the previous session We watched that there could be a sequence where the term could be given by a quadratic formula And in this particular session, we are going to study a sequence Whose nth term can be given by a cubic formula or cubic function or cubic polynomial in n Okay, so the sequence is one three eight eighteen thirty five and sixty one in the First appearance it appears to be having having no trend at all But then on a closer look you will be able to find out that If we go with it, you know the the process of differences, right? That's what we have been doing so far so three minus one is two and then the next is eight minus three is five and 18 minus 8 is 10 and if you see 35 minus 18 is 17 and 35 61 minus 35 is 26 and so on and so forth this we call as level one difference level world difference level one difference then Obviously here also the level one difference is not constant. So hence we can't say it is TN is definitely not linear not linear right So we can't have a linear expression for TN because the first level difference is not constant Let's try to analyze the second level difference Right, so difference of difference. So if you see what is the difference now? three What is the difference here? five and here seven and Here it is nine now here also there is definitely a trend all odd numbers are appearing to be there But then definitely they are not constant. Never mind. Let's go and try third level difference So if you see third level difference, this will be to this will be again to and This is also going to be too, right? So this is where we get a constant difference constant difference and here is where we will stop and we'll say that in such cases where the third level of Difference is constant the nth term will be given by nth term will be simply a n cubed plus bn squared plus Cn plus D where n is our integer which denotes the index and is index or position number You already know position or nth term and is the position of the nth term right and is position right and a b and c and d are a b c and d are mostly rational numbers Rational numbers Okay, this is what we have seen How do we do do we know that is a separate discussion that proof part? We are not going to discuss in this session. We are only going to discuss the implementation part So let's say we have such kind of a sequence. Can I find out a b c and d? So that I Can get an nth term right and it's term expression So let's try and do the same exercise which we did last time So let's say n is equal to 1 So if n is equal to 1 I have from the sequence value as 1 first term is 1 and from the formula So I will write from the formula here and Here I will write the value from the sequence here Okay, so from the formula if you apply n equals to 1 you will get simply a plus b Plus c plus d because all n is 1 and here I'm getting 1 so these are same Let us call this as the equation number 1 When n is 2 Let's put n is 2. So what will happen to the formula formula will become 8 a plus 4 b Plus 2 c plus d isn't it? And from the sequence we will get the value as 3 So let me write the value as 3 here. This is equation number 2. Let's say then Third one, let's go for the third one. So if you see the third value is 8 So let me deploy that value 8 here from the sequence And what will be The formula formula will be 9 Or rather not 9 27 3 cube is 27 27 a n is equal to 3. I am putting so it will become 9 b and And 3 c and b right and this is equal to 8 And the final one Final one is the 18 if you see this is 18 here 18 so let me put 18 over here so this is 18 And this one will become now n is equal to 4. So if you put n is equal to 4 you'll get 64 a plus 4 square that is 16 b Plus 4 c plus d and let us say this is equation number 4 Do check the equation so that you know, you don't make any mistake So that because this is this is going to be a complicated solution. So hence we should keep track of all the Values and equations properly, right? So now we have got four equations and four variables a b and c and d the objective is to find a b c and d So we know solution skills. So hence what do we do? Let's do 2 minus 1 that will eliminate d So if you do that and I will write directly so 8 a minus a is 7 a Then 4 b minus b is 3 b Then 2 c minus c is c and d minus d is 0 and in the rhs you will get 2 Let us call this as equation number 5. This is equation number 5, correct? So 8 a minus a just check 3 b minus b 2 c minus c and d minus d is 0. This is equation number 5 Then you do 3 minus 2 Why the idea is to eliminate d from all so hence you will say 27 a minus 8 that is 19 a Plus 9 b minus 4 b is 5 b I hope we are doing correctly. So just keep keep a check Yeah, 9 b minus 4 is 5 b Then this is c and this one will be 5 equation number 6 and then do 4 minus 3 So what will happen here? So 64. Sorry. I have missed a here. So 64 minus 27. That is 37, right? 37 a just check Yeah Correct 37 a and then 16 minus 9 is 7 b And this one is simply c and 18 minus 8 is 10. So this is 7 Correct now from here. Can we not eliminate c? Yes very much how? 6 minus 5 if I do So c will get eliminated. I will get 19 minus 7 is 12 a Correct and 5 b minus 3 b is 2 b and this one is 3 Let's say equation number too many equations, isn't it? But don't worry. It is going to be easy, right 8 now, let's do 7 Minus 6 that will again eliminate c So 37 minus 19 is 18 a Right and 7 Minus 5 b is 2 b and then minus 5 will give you 5. This is equation number 9 And now I can eliminate b from here. So if you do 9 minus 8 Is equal to then what will happen 18 minus 12 a is 6 a And 2 b 2 b gets cancelled 5 minus 3 is 2. So hence we get a is 1 upon 3 2 by 6 is 1 upon 3 So a little bit more lengthy calculation, but we could solve a is equal to 1 by 3 So when a is 1 by 3 use any equation from 8, let's say let's use 8 from 8 and 9 you can use anyone So let's say 8 So you see this is 12 a Plus 2 b is 3 Is it it? So can I not find so what is uh, can I not find b from here? Yes? I can I can deploy a value a was 1 upon 3 Plus 2 b is equal to 3 So this becomes 4 So 2 b clearly is 3 minus 4 Which is minus 1 so b clearly comes out to be minus 1 by 2 Correct if if there is no let's go back to c. So how to find out c C can be figured out from let's say equation number 5. So let me write it here from 5 Let's deploy value. So 7 into a a was 1 upon 3 Plus 3 b so 3 into minus 1 by 2 Plus c will give you 2 right 7 a Into 3 b plus c is giving you 2. So this is 7 upon 3 And this is minus 3 upon 2 Plus c is equal to 2 Is it it so This will mean c will be equal to 2 minus 7 by 3 plus 3 by 2 Hope the calculations are perfect. So hence if you see this is 6 Then take the LCM so LCM is 6 and then this becomes 12 minus 14 correct and then 9 Okay, so if you solve this you will get 12 plus 9 is 21 minus 14 is 7 correct, so 7 upon 6 Yep, 7 upon 6 is c. So when we get c how to find out d now. So d is There was d d was a plus b plus c plus d was 1. So from 1 you can find out d so from 1 from 1 we had a plus b plus c plus d as 1 Right, so d will be simply equal to 1 minus a minus b minus c. So let's deploy values 1 minus a a was 1 by 3. So 1 upon 3 1 upon 3 Minus d was minus half. So it will become plus half minus c. So 7 upon 6 Let's find the final value. So 6 and this one becomes 6 minus 2 And then to 3 so plus 3 And then simply minus 7 which is equal to 0 Right, so d is 0. So what do I get guys? We get the expression for n term and nth term. It was a n cubed Plus b n squared plus c n Plus d So let's find and you know deploy the value. So a was 1 by 3. So n cubed by 3 D was half minus half. So n squared by 2 And c was 7 upon 6. So 7 n by 6 is the and d was 0. So this is the final relationship. Let's check Let's check whether it is true or not. So if n equals to 1 if I put T1 will be simply 1 upon 3 Minus 1 upon 2 plus 7 upon 6 Which is equal to if you take if you take the LCM like that So 2 minus 3 plus 7 which is equal to 7 9 minus 6 by 6 that is 1 So for T equals to 1 it works, right? If you remember our first term of the sequence was Actually 1 so it works, right? Let's see if T2 gives me 3 from this equation Then our relation will be correct. So T2, T2 will be equal to again deploy 2 cubed By 3 n is 2. So minus 2 square by 2 Plus 7 into 2 by 6. So let's do the LCM again. So 6 And this is 3 times 2. So 8 16 minus 3 12 12 plus 14 So this is 30 minus 12 18 by 6 which is 3 so it works for n is equal to 2 as well So in our probability it is going to work for n is equal to 3 and is equal to 4 n is equal to 5 and so on and so forth. So what is the conclusion of this? Exercise conclusion is that for such kind of Sequences where the third layer of difference is constant It will be expressed by a cubic polynomial And in this case we got the cubic polynomial as n cube by 3 Minus n square by 2 Plus 7 by 6 n right if someone asks us to find out 100th term in this So you do need to just deploy n is equal to 100 and you'll get t100 If someone asks you to find out 500th term so deploy n is equal to 500 And get 500th term like that. So this is what we learned in this session