Telescoping - MATH1048

Added: 2026-05-07

[00:00:01]Yo guys, this is the first video on telescoping.

[00:00:06]So the first thing you need to know in telescoping is the general form where you have one equation and a second equation where they follow each other. Sorry, this is one. So they follow each other. This is R. This is R + one. So they are right after each other. If that's the case, if that's the case, you can apply telescoping.

[00:00:37]So when you apply telescoping, we first distribute this summation sign onto the two of these equations. So the first thing we have is r = 1 n f r minus sum r = 1 n f r + 1.

[00:00:57]We distributed the summation sign or sigma.

[00:01:03]The next thing to do is we open this up. We spread it open.

[00:01:11]The first term is going to be r= 1. So if you put r= 1 there, you're going to have f of one.

[00:01:20]Second term is going to be f of two.

[00:01:25]Next going to be f of three and you're going to have more plus let's try to see the ones at the end. You're going to have f of n minus 2, right?

[00:01:38]Let's add a one to that. f of n minus one. Let's add another one to that. f of n.

[00:01:48]This is this one. and it has to end at n because our upper bound is n and this is f at r. Let's do the second one.

[00:02:02]The second one, the first term is not going to be f at one because if we put one here, it's going to be 1 + 1, which is two.

[00:02:13]So, it's going to start at two.

[00:02:16]The negative sign for that negative sign, it's going to start at two. We're going to have f of two plus f of three plus more plus f at n minus 2.

[00:02:32]Let's check now. n + one.

[00:02:35]It's going to be f at nus one plus f at n plus f at n + one. Right? Those are our last terms. Why is it n + one here?

[00:02:49]Because if you put the upper bound in the place of r you get n + one.

[00:02:57]Okay. What's happening here?

[00:03:00]This f of two is the same as this f of two. This f3 is the same as if three.

[00:03:05]That's the same as that. That's the same as that. That's the same as that.

[00:03:09]What do we have left?

[00:03:12]This guy and that guy. because all of these guys are going to cancel out because of this negative sign right here.

[00:03:22]So, this is positive and this is negative. That's positive and that's negative.

[00:03:28]That's positive and that's negative.

[00:03:29]That's positive and that's negative. And they're all equal.

[00:03:33]That means they have to cancel out.

[00:03:36]That means we end up with f at 1 minus f at n + one.

[00:03:47]So that's the general form for telescoping.

[00:03:51]You have a series that has terms that are following each other.

[00:03:57]You open it up and you cancel all the terms that cancel out. You end up with the first term and the last term of the series. And that's your final answer.

[00:04:06]That's usually how it goes. Sometimes though, you can use another method that I'm going to show you guys for doing this step because sometimes it's annoying to write everything out. So, we can use changing the indexing to get the thing that we want, which is what one of your lecturers uses, which is what one of your lecturer does.

[00:04:29]Yeah, I don't know what I'm saying there. Which is what Tomo does. Let me just say that. um the lecturer for algebra, right?

[00:04:40]But I'm going to show you guys with questions. Now, this is a past paper question. It's question 14 of the problems bank I sent on the group.

[00:04:51]It was in one of the past papers. So, let's do this one. Applying the same rules we applied there. So if we apply the same rules, we have to first distribute the summation sign.

[00:05:04]Let's distribute it.

[00:05:07]That's going to be sum up to n from r = 1 of r + 1^ 2 minus sum up to n from r = 1 r 2. Okay.

[00:05:25]The question is going to tell you guys to use telescoping.

[00:05:31]But if it doesn't, you can tell you have to use telescoping because there's a r + one here and there's a r here. Which means these two terms are following each other and they are the same cuz this is squared and this is squared.

[00:05:47]So you have to use telescoping here cuz it's the same series but this is following that.

[00:05:54]So let's try to open it up.

[00:05:58]Let's see the first term of this guy.

[00:06:02]First term of this guy will be r is equal to 1 and that will be two. So that will be 2^ 2.

[00:06:11]Next will be two here that gives us three. So that's 3 squ.

[00:06:18]Next will be three there. That gives us 4^ 2 and more plus let's check n minus 2. If it's n minus 2 there we have an n minus 2 + 1. This one.

[00:06:41]Okay. Let me show you guys what I'm doing. I'm saying if we substitute into r + 1 all 2. If we substute n - 2 here we have n - 2 + 1 all 2 which is n - one all 2.

[00:06:59]So that's like the second to last term.

[00:07:02]Third to last term I think. So n minus 2 n minus one. I mean you don't have to be specific about these. You can just start with n minus 2 if you want and just go to finish where n is is substituted here and then you'll have the same thing. But I'm going to start with n minus one.

[00:07:22]Okay, starting with n minus one. You don't have to start there. You can start with n minus 2 all squ plus this n minus what's going to follow this. Just add a one here. It's going to be n only.

[00:07:41]The next term, you have to add a one here. If you add a one here, this this going to cancel with this one, and you're going to end up with n only.

[00:07:50]And the last term, if you substitute n here exactly as it is, cuz that's the last term. It's going to be n + one.

[00:08:00]If you don't understand yet, don't worry. You just have to watch the video, watch me do this question and then try to do it after. If you don't understand, you're going to understand over the next questions that you're going to do. You don't have to try to understand everything at once. You just have to see a lot of explanations of the same thing and try to do the questions. Okay, that's the first series where n= 1, n= 2, n= 3. N I mean r is equal to those things. R is N - 2, R is N - one, R is N.

[00:08:37]Let's do the second one. R is one. Well, R is one. We have 1 squ.

[00:08:43]We need to have a negative in front because this whole thing is negative. We have 1 2.

[00:08:50]Then when R is 2, 2. RS3 3 2 sorry RS4 4 2 plus dot dot dot plus by the end we'll have something similar where r is n minus okay let's start n minus 2 where r is n minus 2 just have n minus 2 here squ plus where r is n -1 will just be n - cuz we just substitute everything exactly as it is here. It doesn't change this. This one doesn't change. It's not complicated as this one squared plus the last term where r is just n. So I have n^2.

[00:09:38]Okay. So now you have these two things.

[00:09:41]They're doing essentially the same thing that was being done by the equation on the other page.

[00:09:47]This is a negative sign. It's going to multiply into all of this. If the negative sign multiplies into all of this, the 2^ 2 and the -2 are going to cancel out. The 3^ 2 and the -3 squar are going to cancel out. The 4^ 2 and the 42 are going to cancel out. So all of the things inside here are going to cancel out except for the first one here and the last one there.

[00:10:16]Cuz you see this ends it in. This ends at n and this has an n there. So it can only cancel the terms before and this starts at one. This starts at two. So this can only cancel terms from two onwards. It can't cancel one.

[00:10:36]So what are we left with here? We had an n + 1 2.

[00:10:44]Here we have a minus1 cuz - 1^ 2 is still minus one. That's essentially our answer for this question.

[00:10:56]Okay. So now I'm going to try to do the same question but I'm going to use a different method. I'm going to try to use do the same question but use a different method. Maybe it might be easier for for others to understand that way but I think this one is easy.

[00:11:11]So sum from r = 1 to n r + 1^ 2 - r 2.

[00:11:20]First do the same first step which is splitting it up. So we have sum from r = 1 up to n r + 1 2 - sum of r = 1. I'm just copying this basically.

[00:11:43]Right.

[00:11:45]The first step of this is to understand that you need to have these looking similar.

[00:11:53]You need to have these looking similar.

[00:11:55]If you can make them look similar, you can easily cancel them out. But now they don't look similar.

[00:12:02]Whereas here we saw that the 2^ 2 and the 2 squ looked exactly the same. The 3 squ and 3 squ looked exactly the same except with different signs. That's why they canceled out. We need to try to make that obvious in this notation.

[00:12:19]So the first thing I'm going to do is I'm going to say let s= r + 1.

[00:12:33]If s= r + 1 and I know that here I have an r = 1. Okay, wait. Let me show you why I'm saying s= r + 1. I want to substitute one value here.

[00:12:50]I want that value to be s because if I put that value here that means I have s squ and here I have r squ which looks very similar. s is a dummy variable and r is a dummy variable. So since I'm going to be substituting numbers here substituting 1 2 3 and so on. S would be exactly the same as r 2.

[00:13:13]That's why I'm trying to get this R + one to be defined as S.

[00:13:22]If that's the case, those are my cars outside, by the way. Bought those.

[00:13:30]Yeah, my friends drive that one. All right. So, the R is one. R is one from this.

[00:13:37]R is one. Starts at one. I want to change this to be written or indexed in terms of S. So if I want to do that, S is equal to R + 1. So if R is 1, that means I can put one here.

[00:13:54]So S would be 1 + 1 and S would be two, right? So S is two since R is starts at one. So S starts at two.

[00:14:10]So if S starts at two, that means we have S squ here and S starts at two.

[00:14:15]If S starts at two, our upper bound has to change cuz we've essentially shifted this R by one unit forward and it became s= 2. Our n also has to shift by one unit forward.

[00:14:33]It has to be n + one.

[00:14:36]That's why our n is going to change and become n + one.

[00:14:44]Now let's try to write this using this new indexing format.

[00:14:50]So we have sum the r became s = 2 the n became n + 1 the r + 1 became s squared now minus sum up to n from r = 1 of r 2 that's just this repeated exactly as it is.

[00:15:16]Can you see that these look very similar now?

[00:15:20]It's just that this has term that's not in here and this has a term that's not in here. Let's try to find that out. But first, I want you to confirm that this equation is the same as this. Let's try to confirm that that these two sums are the same. The first value here would be at s= 2. So you put two here. That's 2^ 2. That's the first value. It's four.

[00:15:46]The first value here at r = 1 you put one here that's 2 that's 2 ^ 2 which is 4. So the first values are the same.

[00:15:57]Let's check the second values.

[00:16:00]If you put three now here which is the next value you have 3^ 2 which is 9.

[00:16:08]You put the next value for this one would be r = 2.

[00:16:14]So you put two there that's 2 + 1 which is 3 3 2 is 9 which is the same as this one.

[00:16:23]So these two equations are the same.

[00:16:25]These two series are the same.

[00:16:30]This ends at n + one. Let's just confirm the end. So you just put n + one here at the end and you have n + 1 all squared.

[00:16:39]For this one you put n here in the place of r. You have n + 1 all squared. So they are identical. They are the same.

[00:16:48]You haven't changed anything. This is exactly the same as that.

[00:16:53]Let's see what we can do with this.

[00:16:57]This starts at two, but this starts at one.

[00:17:02]So that means this has an extra term at the beginning where r equals 1.

[00:17:09]Because as soon as r equals 2 that means this and this are the same going forward.

[00:17:14]So let's get that term out. We are going to get that term out.

[00:17:21]r = 1. We have 1 squ and the negative sign here at the front.

[00:17:28]That's that term. The term at the beginning of this one at r = 2. We have 2^ 2. And it's the same as s = 2 because we have 2 2 s = 3 3 2 r = 3 3 2 and so on. Now let's check the ends. The ends we have n + one we have n. I'm going slow. If you think I'm going too slow, you can skip along and try to get to the answer and get to the other questions.

[00:17:58]But I'm going to go fast as we go to the other questions. I'm trying to make sure everyone understands.

[00:18:03]Okay. So this one is at the end and it has n + one. It ends at n + one. But this one ends at n.

[00:18:13]So this one has an extra term at n + one that this one doesn't have.

[00:18:21]Let's get that term out. The term is n + one. And the terms are all squared. So that means it's squared.

[00:18:32]That's that first the last term of this guy.

[00:18:36]Let's write the remainder of these guys.

[00:18:38]Now, since we've taken out the first term of this guy and the last term of this guy, we remove the last term of this guy. So, everything is the same except it ends at n now because we've taken out the n + one.

[00:18:53]S = 2 s 2 minus this guy.

[00:18:59]Everything is the same except for the R. We've taken out the one, so it doesn't start at one anymore.

[00:19:06]Starts at two.

[00:19:09]R equals 2. So if you like, you can do this instead so that it makes a bit more sense.

[00:19:16]We have the one from here and then we start after the one which is two. So this series is still the same as that. This series is still the same as that.

[00:19:30]Now, this is negative.

[00:19:32]This is positive. And they are exactly the same cuz I told you it doesn't matter if it's R or S. As long as we start at two, we end at N. And the series are identical. So, these cancel out cuz you're subtracting them. Which means we end up with n + one or 2 minus 1.

[00:19:56]quite slightly longer actually. I think I'm going to do the next questions using this method for you guys' sake. Um maybe do at least one in this method again.

[00:20:09]But this seems very short.

[00:20:23]This is question 18 of the questions I sent on the problems bank on the group.

[00:20:32]Try to do this question and then come back to confirm if your answer is correct. You can pause the video, try to do it and then come back to confirm if your answer is correct. So I'm going to be a bit faster on this one.

[00:20:44]We have we distribute this guy. The sum.

[00:20:49]So we have sum up to n r = 1 of 1 / r + 1 minus sum up to n of r = 1 / r + 2.

[00:21:04]Right? It's the same as this just distributed.

[00:21:08]Now I'm going to open it up.

[00:21:13]First series is going to be from r is one. It's going to be 1 / 2 plus with RS2 1 over 3 RS4 1 over 4 IO.

[00:21:29]You get the idea.

[00:21:32]Plus by the end we are definitely going to have an n minus 2 somewhere 1 / n minus3 somewhere 1 over oh no my bad minus one it's been a while since I've done these 1 / n and then a 1 / n + one Right.

[00:22:04]Now let's do the second one. Negative into 1 / if you put one here, you get a three + 1 / 4 + 1 / 5 + dot dot dot + 1 / n - 2 + 1 / n -1.

[00:22:25]Right? I'm just going to the next term.

[00:22:27]So the next term I'm just going to add a one here. Next term I'm going to add a one to the m.

[00:22:35]Next term I'm going to add a one again.

[00:22:42]It's still not the same as this if you put an n here. So I'm going to add another one.

[00:22:50]Right now it is the same as this because if you put an n here it's just n plus two. You do have to end at n plus2. that that's what that means.

[00:22:59]Now, from what we have here, you can tell that the 1 over 3 is going to cancel the 1 over three. The one over four is going to cancel the one over four, and so on. You don't scratch the numbers out when you are like writing on your exam.

[00:23:12]You don't have to do that.

[00:23:16]Uh, and then that's going to cancel with that. This going to cancel with that.

[00:23:19]This going to cancel with that. I'm just scratching it so you guys see what's happening.

[00:23:24]Okay. So what do we have left? The first guy and the last guy.

[00:23:30]So that means we have a half - 1 / n + 2.

[00:23:38]That's your answer. That's your full answer for the question.

[00:23:41]Straightforward.

[00:23:43]I'd say this is slightly longer. So if you guys are complaining about telescoping, I think this is what you guys are talking about.

[00:23:51]Um, if you are fine with the second method, just skip along the video. But I'm going to do the second method here. Try to move a bit quicker on it.

[00:24:00]So, we have some So, you could have done it this way or you could have done it this way. Not both. You're going to waste your time on the exam. R = 1 N 1 by R + 1 - sum N R = 1 1 by R + 2 right this is this step right with the summation sign distributed what do we want we want these two guys to be similar how do we make them similar you have to make this look the same as this.

[00:24:40]So how do we do that? Check this guy.

[00:24:44]r + 2. It can be written as r I mean r + 1 + 1.

[00:24:54]So if we can get a way to put this r + 1 to be one term, we can get a something + one. Something plus one, which is the same as this something plus one.

[00:25:06]So let's let Q equals R + one. Doesn't matter what alphabet you use. It's a dummy variable.

[00:25:19]Q equals this. Right?

[00:25:23]Now we need to change these bounds as well to be in terms of Q.

[00:25:28]We had R = 1 here. That means our Q has to be 1. This one + one which is two.

[00:25:43]And our N has to change. We moved our our lower bound by one unit forward.

[00:25:50]It's now two. It was one.

[00:25:53]So our n has to move by one unit forward too. It has to be n + one.

[00:26:00]Now let's write the full things.

[00:26:03]We have sum from n of up to n from r = 1 of 1 / r + 1 minus sum 2 n + 1. Our upper bound has changed from q = 2.

[00:26:24]1 / what is r + 2? It's Q + 1, right?

[00:26:33]R + 2 is Q + 1. Um, I don't know if I made that here here, but that's what we are getting to through all of this cuz r equals Q = R + 1.

[00:26:49]It's equal to this guy.

[00:26:53]So that's that becomes Q + 1 because it's the same as that guy.

[00:27:01]Now we have this. If there's like things I don't explain clearly and you guys think I should explain them more clearly, tell me on like WhatsApp the group. If there's things I explain too much that I shouldn't also say that on the group. Okay, this is what we have now. This starts at two. That means this has an extra term at the beginning where r equals 1. Let's take that term out.

[00:27:25]Put r = 1 here. You get 1 / two. That's 1 / 2 plus then the remainder of the sum is r = 2. Right? Now starting at two cuz you've removed the first term. n 1 / r + 1 minus check this one.

[00:27:48]This one is starting at the right number two, but it's ending at n + one, whereas this one is ending at n. So, we need to deal with the n plus one term. Let's deal with that. We're going to have the last term.

[00:28:05]That's going to be the last term.

[00:28:07]So, that's one over n + one. In the state of q, we're going to put n + 1 + 1.

[00:28:17]That's that last term.

[00:28:20]And then what's left over after we've removed the last term.

[00:28:25]What's going to be left over is sum up to n because we've removed the last one which was n + 1. Q = 2 1 / q + 1.

[00:28:37]Right? This is what we have.

[00:28:43]So we have the half there plus this guy sum r = 2 to n 1 / r + 1 distribute the negative sign you have negative sum to n from q = 2 of 1 / q + 1 minus this negative sign goes there as 1 / n + 2.

[00:29:16]Right? Now, what do we have here? We have a half.

[00:29:20]This cancels with that cuz it's exactly the same. It's just that this is Q, this is R. This is R, this is Q. But again, I said these are dummy variables. So if the bounds are the same 2 and n 2 and n and the equations are the same 1 / r + 1 1 / r or 1 / q + 1 that means these two series are the same so they cancel out we have a -1 / n + 2 which is what we already had here.

[00:29:56]Okay. A good rule to check if you're correct is to just come back to the equation.

[00:30:02]Check h this one ends at r at n + 2, right? So it's going to be 1 / n + 2.

[00:30:15]So this one ends higher than this one because this one ends at n + one.

[00:30:22]So this guy's last term has to be in the answer 1 / n + 2 and it has to be negative cuz this guy is negative.

[00:30:31]So you take the last term of the larger guy and the first term of the smaller guy. Smaller guy's first term is going to be 1 + 1 which is 1 / two. This is going to be 1 / two.

[00:30:42]That's why we have the one over two. So essentially you can get the final answer before you even do the question but you have to do these steps to show your marker that you are a student who is respecting the rules and following doing questions and stuff. All right, this is the next question.

[00:31:02]This one is not on the document I sent because it's not a past paper question.

[00:31:08]I created this one just to show you guys how to deal with linds before I get to a slightly more complicated one for lens.

[00:31:15]Okay, let's do this one.

[00:31:17]For logarithms, you apply the logarithmic laws.

[00:31:22]Remember, there's a logarithmic law that says log of a over b is equal to log of a minus log of b.

[00:31:39]This is the log law we have to use because remember on all of the telescoping things we've done it has to be a negative. There has to be a negative somewhere. We need to have a different sum or else we can't telescope it.

[00:31:56]So let's try to apply this. The logarithmic laws can give us the negative.

[00:32:02]So we have sum from r = 1 up to n of lin r + 1 / r.

[00:32:11]We apply this logarithmic law denominator and numerator denominator and numerator denominator numerator.

[00:32:20]That means we have sum r = 1 up to n of lin r + 1 at the top minus lin r That's what we have. Now we have our negative with smile.

[00:32:37]Now we distribute this guy.

[00:32:41]If we distribute that guy, we get sum from r = 1 up to n of lin r + 1 minus sum from r = 1 up to n of lin r.

[00:32:56]This is what we have. I'm only going to do the short way for this one, not the long way.

[00:33:01]If you guys want the long way explained more, of course, say on WhatsApp.

[00:33:05]Anything you need, just say on WhatsApp and we'll see if you can get it done.

[00:33:10]So, the short way you open it up, you have Let's check the answer first in fact before we even start from this. This ends at r + one.

[00:33:24]This ends at r. So, this is larger.

[00:33:29]So, If we put n here, our last term has to have an n + one here. Last term is going to have a lin n + one.

[00:33:42]Our first term because this starts at r + one but this starts at r. So this has a first term that this guy doesn't have.

[00:33:51]First term is where r is equal to one.

[00:33:54]So that will be lin one.

[00:33:56]Lin one is zero.

[00:34:00]So basically this is what it simplifies to. Our final answer has to be this.

[00:34:07]Let's continue here. Let's try the first series. Put one here. You get lin 2.

[00:34:15]Put two you get lin three l four plus dot dot dot plus I'm going to go lin n minus one lin n l n + one minus the second series lin 1 + len 2 + lin 3 + lin 4 beautiful in writing I know plus dot dot dot plus lin n - one + lin n this one ends at n because it's just lin of the value and the last value is Then we distribute the negative sign we get that this becomes negative. This becomes negative and all of these become negative. So this cancels that guy.

[00:35:22]Remember don't cancel on your test. I'm just trying to show you guys what happens. I'm not going to cancel on the next question.

[00:35:32]Okay. So what do we have left? That guy and this guy.

[00:35:38]So basically we have lin n + one minus lin you get the idea one.

[00:35:50]How do we find lin one? Because I just told you that when I was trying to get this one what value it is.

[00:35:58]Um let's try take a bit of a detour here. If you know what lin one is and know why you can just skip ahead.

[00:36:06]Lin one. What does this mean? This means e to some exponent that I don't know, let me name it k is equal to one. That's the that's the definition of this. So it means this is basically k.

[00:36:24]So k is this. Therefore, what should k be? What exponent would I have to put e to for me to get one as my answer?

[00:36:36]So we know that anything to the power of zero gives us one. So that means k equals 0 because e to the 0 = 1.

[00:36:50]That's why 1 is equal to z. log lin whatever logarithmic base you have log of one is equal to z. So our answer is lin n + one.

[00:37:04]That means this whole thing is the same as lin n + bus one.

[00:37:13]Okay, let's go to another one.

[00:37:20]Let me help you out.

[00:37:23]The only thing that's going to change here compared to the other one is that now you're going to take the two to the front because that's another log law.

[00:37:32]the log law where you take the exponent and make it a coefficient.

[00:37:37]And then basically you're going to apply the same thing and get to the answer.

[00:37:44]But here you have a 99. You don't have an N. So you have to deal with the 99 instead of the N. So try to pause the video and try to do this question.

[00:37:59]Okay. I hope you tried it. If you didn't um that is um also existence sum from r = 1 to 99.

[00:38:14]This two we take it as a coefficient log base 10 of r / r + 1. Right? That's a log log. You can always take exponents of the thing that is inside and put them as coefficients.

[00:38:30]Now, if you have a two here and a sum, you can always take the two out. You can always take a coefficient or a constant outside of the sum. That's another law of um sums. It's a law of sums now or sigma notation. You use that you get the two to the front.

[00:38:51]So, we have everything inside.

[00:38:53]We have sum r = 1 99 log base 10 r + 1.

[00:39:03]That's what we have. Let's go on two.

[00:39:08]Now let's apply the log law we had applied. That's like this law.

[00:39:16]That means here we'll have sum r = 1 up to 99 of log base r minus sum from r = 1 up to 99 of log r + 1 we set then close the bracket.

[00:39:43]Okay, once again, I'm going to do the short method. If you want a longer way, this one, tell me on WhatsApp and I'll see if I can make a plan.

[00:39:57]I will probably make like a short video explaining it again with like two problems or something like that. I don't want to make anything you don't need though. Okay, so this is what we have.

[00:40:10]So, we have two as the front guy.

[00:40:14]Let's open these series up.

[00:40:19]If we put one here, that's r= 1 cuz it starts at 1. We're going to have log one, which is zero.

[00:40:30]If we put two here, we're going to have log two. So that's log base 10 of two.

[00:40:39]Then we have log base 10 of three and more logs plus near the end we have log base 10 of n minus 1 plus log base 10 of n.

[00:40:59]It ends at oh not n. My bad. My bad.

[00:41:02]This time we know that n we know that n is 99.

[00:41:07]Should I force it? Nope. You guys get the idea. Here we have 99. That's our last term. It's 99.

[00:41:16]The term here 98 cuz we keep continuing. So 98 before that will be 97 and so on.

[00:41:24]Then we have a negative into and then we write this one where r is one. We have log base 10 of 2. RS2 log base 10 of 3 plus dot dot dot + log base 10 of 98 plus log base 10 of 99.

[00:41:55]This one has r + one. So the last term will be 99 + one inside of the log. That's log base 10 of 100 cuz that's what 99 + 1 is.

[00:42:10]It's 100.

[00:42:14]What happens here? This cancels that.

[00:42:17]This cancels that. All these bad boys cancel out in the middle until you have this bad boy and that bad boy.

[00:42:26]Which means you have two into that's zero. It doesn't matter. You have negative log base 10 of 100 which means you have -2 log base 10 of 100. Okay. Now I want to show you the answer here is two is two for this guy and the answer is four. If you get that, the answer is -4. The final answer is4. Just skip a few seconds ahead, but I want to help the people who can't figure out what this is.

[00:43:06]So, k equals log base 10 of 100.

[00:43:14]What is this saying? It's saying 10 to some exponent gives us 100.

[00:43:25]What is the exponent? That's what this question is asking. That's what log is asking. What is the exponent we have to raise 10 to to get 100?

[00:43:38]So what exponent do we have to raise 10 to? 10^ squ is 100. So that means k is 2.

[00:43:50]That's why here we have -2 into 2 which gives us -4.

[00:43:58]Right? That's what we have as our final answer to this whole question.

[00:44:07]Now, we've got to do this one. All of these are carefully selected to be different so that you guys learn as much as possible. This one is also not in your your problems bank, but this one is in question 13.

[00:44:22]It was in a pass paper. That means this was in a pass paper.

[00:44:27]This one was not. This one was okay.

[00:44:31]Let's come to this one. Actually, this one is seemingly I'm just realizing now.

[00:44:40]It's the same as this. Crazy.

[00:44:43]But I'm going to show you why it's the same. First, if you have something like this and they say use telescoping, you have to remember your partial fractions.

[00:44:51]first block. So let's try to apply the partial fractions. How do we find the partial fractions of this? Let's first deal with this alone. Forget the summation sign for now. You have 1 / r + 1 r + 2.

[00:45:09]Okay, if that's the case, we can write this as a over r + 1 + b over r + 2.

[00:45:20]They told us to use telescoping. That's why we applying this because we know in telescoping we have to have two things that are being subtracted and the two things should follow each other. So these are following each other cuz this is r + one. Right after r + 1 you should have r + 2.

[00:45:39]Now let's find it. That means we can write this as a into r + 2 + b into r + 1 all / r + 1 r + 2.

[00:45:56]Okay, now we just need to equate this to one.

[00:46:00]So a into r + 2 + b into r + 1 = 1.

[00:46:11]Multiply out the B's into the inside.

[00:46:13]You have the B's and the A's. A R plus 2 A plus B R. If you know how to find the partial fractions, if you know how to find the partial fractions, you can find the partial fractions. If you do that, the equation will turn exactly into this one. I think if it doesn't, then we'll see. But I think it turns into this one.

[00:46:36]If it does turn into this one, that means you already know how to find the final answer. So, you can attempt it if you know how to find partial fractions.

[00:46:43]You just find the partial fractions and then do it exactly how we did this one.

[00:46:50]Plus B equals 1. This video is becoming a bit long, but it's fine.

[00:46:56]Group these guys. Take out the R. So you have r into a + b + 2 a + b = 1.

[00:47:08]We don't have an r here.

[00:47:11]That means we shouldn't have an r here.

[00:47:13]That means these need to be zero to to cancel out this r. So a + b equals z, which means a is equal to minus b.

[00:47:27]Now let's get this guy. If these are zero that means 2 a + b = 1.

[00:47:35]2 a + b = 1. a = b 2 = - b. I mean yeah that sometimes my tongue slips. I so I substituted this into the place of a.

[00:47:52]This now becomes - 2 b + b = 1.

[00:48:01]We subtract these. This becomes minus b = 1. Which means b is equal to minus one.

[00:48:10]And a is just a negative version of b.

[00:48:13]So a has to be positive one.

[00:48:20]Now let's go build our thingy. This that means we have 1 / r + 1 r + 2 is equal to a is 1. 1 / r + 1 minus b is - 1. That's why I'm putting the minus 1 / r + 2. Yep.

[00:48:46]If this is the same that means the sum from r = 1 up to n of 1 / r + 1 r + 2 should be equal to the sum from r = 1 up to n of 1 / r + 1 - 1 / r + 2.

[00:49:13]Right? They should be the same.

[00:49:16]which means we can just evaluate this.

[00:49:18]So here the main problem with this question was finding partial fractions which means it's important for you guys to re to revise your block one stuff because this was block one and you didn't learn in block two to find our answer here. I'm not going to do it. Why? Because we did exactly the same thing on the on the other question. What I'm going to do is I'm going to give you the final answer. So I want you guys to try to find the answer to this.

[00:49:46]So the final answer in the way that I showed you guys how to find the final answer.

[00:49:52]This one starts at one here, which means two here, which is 1 / 2.

[00:50:04]This one starts after this, but it ends after this again. So it ends at 1 / you put n there n + 2.

[00:50:19]That should be your your your final answer.

[00:50:24]But don't obviously don't try to skip to this. You have to follow the steps.

[00:50:28]You're a student. I am a second year student. Yes. Yes. I can took this.

[00:50:35]Okay. Let's go on. Now we have this one.

[00:50:40]Okay, this was on your guys' thingy on you guys' pass question papers. It was it's question 17 on my problems bank that I sent you guys on the group. But the first thing is they say show that this is equal to this.

[00:50:56]So they helped you out here on the question that was on the pass paper.

[00:51:00]They tell you to show that this is equal to this. You should already be getting a hint cuz three is after four, four is after three, which means you get to use telescoping.

[00:51:11]So use telescoping on this exactly the same way you would have done here and you get your final answer.

[00:51:19]And then there's an extra question which says find the sum to infinity.

[00:51:27]Please try to do this question now.

[00:51:29]Pause the video, take out your exercise book, and try to do it.

[00:51:38]Okay, because this video is getting long, I'm not going to do it, sadly, cuz I know you guys are not going to watch long videos.

[00:51:50]You know that you just have to prove this thing. So, you don't I don't have to do it. You just have to do it and see if it's the same. And it is going to be the same using partial fractions. in that way we use for this you use the same way we used for these equations but I'm going to give you the answer so the first one since we know the partial fractions are like this the first one has to be you put one here on the smaller one it's going to be one over four you put n on the larger one it's going to be 1 / n + 4.

[00:52:29]That should be your final answer. You guys do it and find this answer.

[00:52:35]Okay, this is what I can help you guys with the sum to infinity. If they say find the sum to infinity of something you've already reduced into something like this, a shorter version of it that doesn't have sigma notation inside of it, this is what you do.

[00:52:54]You say you find the limit as n approaches zero infinity I mean of the sum okay of the partial sum I guess this is called a partial sum because it ends at n it doesn't go to infinity it ends at n so it's part of it so if you find the limit as n approaches infinity of the partial front of the partial sum which is SN. That's what this is. You get the you get the sum to infinity.

[00:53:33]It's equal to the sum to infinity from r = 1 of 1 / r + 3 r + 4. Oh, sorry.

[00:53:44]Okay. Now, let's do it. We have limit as n approaches infinity of 1 / 4 - 1 / n + 4 and this doesn't change. We can distribute the limit sign of course to here and to there. The limit for this guy is just 1 / 4 doesn't change minus the limit to this guy. Limit as n approaches infinity of 1 / n + 4 1 / 4 minus let's check this guy.

[00:54:24]If n approaches infinity, n is growing very big. You can check this on a calculator. If you put like 9,000 here, you get something that's very close to zero.

[00:54:36]put like 90,000 it gets even closer to zero. If you put like one like million or 9 million you get something very very close to zero.

[00:54:47]So if that denominator keeps increasing and the numerator doesn't change the value approaches zero. That's part of the limit laws. So - 0, which means the sum to infinity is 1 over 4.

[00:55:08]Okay, that's what that is.

[00:55:12]Last question.

[00:55:14]It shouldn't be the last question, I guess. Let me let me see which one I should do. I was planning on doing this one, which is question 15, but I think you guys have the capacity to do this one if you've done the ones before, cuz it's essentially the same thing.

[00:55:27]Open it up and try to find your thing. I guess sum to infinity of this of the thingy.

[00:55:34]We can do Yeah, we can do that. Sum to infinity of this. So, let's just try to find the answers to this. It'll be the first one of this guy, which be 1 / e minus with the nth one of this guy 1 / e to the n + one.

[00:55:53]That's this the thing. So the sum to infinity should be the limit as n approaches infinity of 1 / e - 1 / e to the n + 1 limit of this guy e is a constant. It doesn't change. It's a number like four or five or six or anything. So it doesn't change. Just remains that. But this guy the limit as n approaches infinity if we put something here one we get a small number for e to the something but the more we increase n the more we increase this n that means this increases because it's the exponent of e right if we increase the powers of something it gets larger so this thing gets larger Cuz 2^ 2 is sum is 4 but 2 to the^ 3 is 8. So it gets bigger as the exponents increase.

[00:57:03]As it approaches infinity this approaches infinity.

[00:57:07]The denominator is getting smaller. The denominator of this which means this whole thing is approaching zero. Cuz if the denominator is approaching infinity the numerator or the whole value is approaching zero. So that's zero. So this is your final answer.

[00:57:26]That's question 15 on the questions bank. It was on a past question paper definitely cuz I took these directly from past question papers.

[00:57:37]This would have been the last one.

[00:57:41]Yeah, I'm going to do this one.

[00:57:44]So they give us they gave us all of this like stuff um in a lot of words. They wrote a lot more words than I did question 16. I hated that, but who cares?

[00:57:57]Um, if the lecturers can punish us. So, they gave you this. They said the sum of this guy is equal to a n + b to the^ 4 + c. You have to find a, you have to find b. You have to find c.

[00:58:14]But then they also gave you a hint. They said 4 r cubed plus this this this is equal to this.

[00:58:28]Okay, if you don't understand how the question is phrased here cuz I like paraphrased and summarized, go to question 16 of the questions bank on WhatsApp. You're going to see how it's written on the questions bank which is how it was written on the question paper. So essentially this is what we have. They gave us sum of 4 r cub + 6 r 2 + 4 r + 1.

[00:58:54]Oh, I forgot to say r = 1 and n.

[00:58:58]Okay, this whole thing inside is equal to that. And we know that we have to have a negative sign and two things following each other to be able to apply telescoping. So we rather use this version of it. That's sum r to n from r = 1 of r + 1 to 4 - r 4.

[00:59:24]Now we can uh do our first step of telescoping which is distributing this guy.

[00:59:30]So we have sum n r = 1 of r + 1 4th minus sum up to n r = 1 of r 4.

[00:59:44]Now we we open this thing up.

[00:59:47]If r is one, we have 2 to the 4th for this one. Then it's 3 to the 4th. Then it's 5 to the 4th and so on. Plus, if r is one, we have 1 to the 4th. Oh, no, my bad. Start finish. Let's finish this one up first.

[01:00:07]We have n - one to the 4th + n 4th + n + 1 4th has to be n + 1 minus this one. Now 1 2 1 4th + 2 to 4th + 3 4th + 5 4th + dot dot dot + n - 1 4th + n to 4th. This one has to end at n 4th. Can't end at n + one.

[01:00:50]Now this guy cancels. They cancel. They cancel. You're left with this and that.

[01:00:55]which is 1 but but it's negative plus n + 1 to the 4th right this question just forgot to explain this I was trying to confuse you guys it like it had a lot of words and like a lot of stuff about like I don't know history of humanity and stuff but the main thing for you to notice is they they said something about partial sorry they said something about telescoping therefor you use telescoping do what you know like just do what you know um but if you do understand the words um understand the words and do what you are told to do right so let's write this neatly start with the positive version of the thing start with the positive thing I mean to the fourth minus one and they said that the sum of this guy was equal to this thing so and they said we should find A, B and C.

[01:01:56]We have written it in this form now.

[01:02:01]So let's find our A, our B and our C.

[01:02:05]Our A is the coefficient of N.

[01:02:08]That's one.

[01:02:10]So therefore A equals 1.

[01:02:14]Our B is this one.

[01:02:19]So our B is also one. Our C is the left over one. It's negative one.

[01:02:26]So our C is equal to one.

[01:02:33]Right, that's the end of the video. I hope you guys understood stuff. If you need any other things, make sure to vote on the poll so that I can make that video for you guys next time. Good luck on your um um worst um years of your life at VS.