Klausur UNI Mathe – Beweis durch Vollständige Induktion

Hello everyone! Today I would like to show you an example of how complete induction works. We want to show that for all n of natural numbers with n greater than 0, this statement applies here. So that you can always calculate this sum on the left-hand side like this. And since we want to show this statement for all natural numbers greater than 0, the complete induction is perfect here and we just go through it in these three steps. And then we have shown the whole thing. First step, the induction start. It is always very nice to us, because at the induction start, this is what we should prove, we just have to look at what our first n is, i.e. from where the whole thing should start for which this statement should apply. And there it said for all n greater than 0, i.e. the first natural number greater than 0, then n is equal to 1. And we just put 1 for every n that we can find here somewhere, we can all mark them green. For every n we put 1 and see if the equation applies. On the left-hand side we have this sum of k, k is not changed, this is just the index of our sum. But there is a green n, we put 1 there and there is again a k in our sum, we do not replace that, we leave it as it is. But let's see if that results in the right-hand side. For that too, we put 1 for every n. That means we have 1 times 1 plus 1, you can already calculate that, that's 2. 2 times 1 is 2 plus 1 is 3, that's what's in the numerator and the 6 in the denominator. Let's see how we can calculate the sum here on the left-hand side. k is our index, it starts at k equals 1. That means we put 1 for every k that we find here. Then we have 1 raised to the power of 2. And then the sum is already over, because it only runs from k equals 1 to 1. So we only have one sum, nothing else runs. So 1 is on the left-hand side, 1 to the power of 2 we can also simply write as 1. And on the right-hand side, if we calculate that, 2 times 3 is 6, divided by 6 also results in 1. So our induction beginning is correct, the equation works for n equals 1. And that's all you have to do, then we have completed the induction beginning. Second step, the induction prerequisite. It is now even easier than the induction beginning, because we don't have to calculate anything here. We just have to write one thing down. And with the induction prerequisite, the following comes up. We have just found an n in the induction beginning for which this statement is correct. And we just write that down now. There is an n in the natural numbers. If you had a condition for it, you had to add it now. So our n should really be greater than 0, so that the statement that we should show is correct. So that this equation applies here. You can then simply write it down from your assignment sheet. It may seem a bit ridiculous to you, but the induction prerequisite is super important in the further course. So remember it, I'll mark it green here now, it will appear again in a moment. We will have to use it somewhere. But otherwise, just write it down here in this step. And then we have already got the induction prerequisite. Third step, the induction step. This is where the masterpiece happens, because so far there has been nothing. In the induction step, we now have to close from n to n plus 1. That means we first have to look at what is to be shown for us now. We are not supposed to show this statement for n, because that is how it is written here, but for n plus 1. So now we have to use n plus 1 for every n that we find here, so that we can see what we have to show at all. You always do that. Use n plus 1 for every n, that is, on the left side. We leave k as it is, but there is an n, we put n plus 1 there. k is there again, we leave it as it is. And on the right side, there is a little more going on. We replace n with n plus 1. Nice brackets around it, because all of this is supposed to be multiplied with the next bracket. There is an n in there again, we replace it with n plus 1. Then we have to be careful, because at the back there is also a plus 1 that was already there before. You can also add them up. 1 plus 1 then adds up to 2 brackets. And in the next bracket we have 2 times, we replace the green n with n plus 1, so a nice bracket around it. And at the back there is also the plus 1 and in the denominator we have the 6. You can simplify it a bit. We can leave that here, but at the back, the bracket, we can add a little more. 2 times n is 2n and 2 times 1 is 2. And at the back, the plus 1 is added. That's 2 and 1 is 3. So we can simplify the back bracket to 2n plus 3. And then we already know what we have to show now. I have now written that again on the next page. We are still in the induction step. The red is what we still have to show. But the green, you noticed that, that's the induction condition. We can use that from now on. That is, as soon as we come across a sum that looks like this, that goes up to n here, then we can replace this sum with this expression here. So always have this induction condition in mind where we want to go. So we want to create something here that we can replace with this. But we have to get to the red. That means we start on the left side, take this sum from k equals 1 to n plus 1 and see how we can now transform it so that we can apply the induction condition. This sum runs up to n plus 1. That's one too many sumands. We only want to run up to n, but with sums it's not a problem. We can divide them into two sums, just one sum plus a second sum. The first sum starts at k equals 1 and then does not run to n plus 1, but only up to where we want it, only up to n. And the second sum does the rest. The successor of n is n plus 1. And how far should our sum run in total? Up to n plus 1. Then we enter that up here. You don't change what was in the sum. It just comes in again. And now we have already managed to make the part here so that it is exactly our induction condition. And we can now replace that with this part. So mark it, that you apply the induction condition here. And instead of this sum, we now write the part there. So just write this big break off once. That's our induction condition. And it always has to be applied at some point. What we must not forget, there is still a sum at the back. We have to calculate that now, because it only runs from n plus 1 to n plus 1. That's just one sum, so we only have to use n plus 1 for the k here. That means, if we do that, instead of the k we write n plus 1. Then we have to make a bracket around it, because it should still be squared. And then we can just do that. So we did it. It looks very good already. Not perfect after this one, but already more similar than our sum. That's why this is the intermediate step. I have it here again. This is how far we have come so far with our transformations. Now we just have to make sure that we get further. We always have a view of where we want to go. That's what the structure should look like. That's not the case yet. That's just one big break. We have a break here at the moment, plus something else. That's why let's put these two parts together. So we also make a break out of it here with denominator 6. But that only works if we also add a 6 to the numerator. Then we have two breaks with identical denominators. We can add them by adding the numerators. If I move around a bit here, the denominator will only become a 6. So a 6 can go away. And here we also make a big break out of it. And then we've already created something very similar in terms of structure. A big break. And the denominator is already identical. That means we don't have to go to it. But the counter doesn't look like this yet. So we have to see how we can somehow compensate for that. We have three brackets here that are multiplied with each other. We don't have that here yet. We actually have a big sum here. That's not so good yet. We want to get a product. We can do that by clamping things out. And if we look at what's happening here twice, then we have the n plus 1 up here and that's also happening back here. And practically, it's also a factor in the product here. So it's a very good plan to clamp this n plus 1 out of the sum. Then we have a big break out here. We can take over the 6. And we're going to pull out the n plus 1 now. Big break. What's left? Here, the n and the other bracket from the first sum. Then comes the plus. And here in the back, the green pulled out, the 6 remains. And n plus 1 was in there twice. So there's still one n plus 1 left. Big break. If we compare it now, we have now pulled out n plus 1 as a factor. That's very good. But the back part is not quite what it says here yet. You could see if you could clamp something out even further, because we have a sum here again. But that's not the case this time. That's why I would suggest here at this point, let's solve the brackets first and see what comes out of it. So we'll take over the 6. n plus 1 up here is great. We don't touch that. That's exactly what we want. But let's solve here. n times 2n is 2n to the power of 2. n times 1 is n. And here in the back, the bracket is released. 6 times n is 6n. And 6 times 1 is 6. We can just summarize that a little bit. 1n and 6n are a total of 7n. And here, this intermediate step, I have it here again. That's how far we've come so far. Now the last thing we have to show is that this part is nothing more than this part here. And here's a tip for you. Don't try to start from here and factorize the whole thing so that you get this shape, but just start from the back. Start from here and see if it's the same as this one. So if we multiply the brackets here, just for you on a piece of paper. So n times 2n is 2n to the power of 2. n times 3 is 3n. 2 times 2n is 4n. And 2 times 3 is 6. That means if we add 3n and 4n together, we end up with 7n. And that's exactly what's written here. Then just write it down. So you don't have to show me how to get this factored form from here, but we just write it down that this is actually exactly the same as where we want to go. It's the same. We just went backwards and showed it. And then you're done. So don't try to come up with any special art here, but just see if it's the same if you start from the back and write it down exactly the same. Then you can put your box underneath or put your QED underneath. And then we showed the induction step, namely, that's exactly what we should show. We showed. And with that we also showed the complete task. Then I hope it helped you. If you have any questions, feel free to write them in the comments. Otherwise, have a wonderful day and we'll see you in the next video. Take care!