Show that if n and .integers with *ksn,tken () <m/-'...

Show that $n^{2}+n$ is an even number for all integers $n$

Let's start with the right hand side, two to the power and could be rewritten as the quantity of one plus one to the power button, which means we can apply the manual meal. Fear a would equal one B would equal one and end obviously equals. And so, in summation notation, we have the some of K equals zero to end of and shoes K times A to the n minus K powers always one to the end minus K Power times B which is also one to the cape Our you well, we know one raised any positive power is just one. So this is really the summation from zero toe en of n choose k. Well, we can write this then as and choose zero plus And she was one plus and shoes too and so on, all the way until and choose which is the given expression. So there we have proved that this is equal to which two to the power of

Were asked to show that and into your end is prime if and only if the Oiler five function of N is equal to end minus one. So will prove this by contradiction. Now let NB prime and the only factors of n r. One and end The greatest common divisor of n and every non factor of n is one. So we have, for example, the greatest common divisor of one end this one the greatest common divisor of two and then is one all the way up to the greatest common divisor of n minus one in one and then which is one. So it follows that all positive integers strictly less than and are relatively prime to end. So it follows that Oiler five function of N is equal to end minus one. Since there are n minus one positive integers strictly smaller than n for the second part let five an equal and minus one. Now we know the greatest common divisor of n and n is n and we know that there are n minus one positive injures smaller than n. So it follows that every integer smaller than end has to be relatively prime to end and therefore it follows that the greatest common divisor of one end, of course, is one the greatest common divisor of two. And then is one all the way up to the greatest common divisor of N minus one and then, which is one? This means that none of the integers smaller than n and greater than one are divisor is of n therefore, by definition, en is prime.

We will use induction to prove. So we assume that. So zoom assume that for and equal to K. The inequality is true, inequality inequality is to it's true. So that means you have suki to be equal to three kids to be less than or equal to 2 to the Power Key. So now we prove so we prove for and equal to key platz one. So then this implies that you have to K plus one To be less than or equal to shoot the power kids like one. So you have To keep plus one less than or equal to apply in the CCR. You have to use the power key to. So then to council south you have K plus one To be less than or equal to 2 to the power key. We also know that N. Well always uh huh. The Great start than one meaning meaning that's it's it is true. It is true debts. It is true that okay is greater than or equal to one. So then okay lets key. Yeah K plus K will be greater than or equal to one plus K. So that you have to think a to be greater than or equal to one plus key. So what we have done it's so we have so we have proved that two K. Is greater than or equal to one. So using this we can right? You can write this so we can write ski You can write scape plus one to be less than or equal to 2 to the Power Key. So you have suits the two key here to be Less than or equal to keep plus one. And this would be less than or equal to two to the power key To obtain. It implies that Suki. It's less than or equal to 2 to the Power Key. So this is what we wanted to do. So we wanted to prove by induction that, too, and it's less than or equal to two to the power in which we have been able to show here by induction.

Okay. This problem requires a bit of thought. It's different. We're dealing with inequalities, right? Saying that for any natural number And remember, all positive. Introduce that it's going to be a lesson to to the end. So this seems obviously true. Of course, we plug a one in for end, we'll get one is less than two to the first. That's true. Uh, we want to solve by induction. So we'll assume that the K cases true that K is less than two to the cave. All right, then we can look at the K plus one scenario. All right, so we'll have K plus one is less than two to the K plus one. Uh, we can rewrite that right side if we'd like as to to the K times two the generally, you know, we might just get stuck here on this is if if we've just been working a bit mindlessly, you might really be trying to get that factor of two, and we're not interested in that. Generally, and inequalities were trying to solve in this strategy, they're saying, Okay, some number end, that's less than why. All right, And then our goal then, is to maybe find a number X. That's a lesson end so that afterwards we can say, Well, excess an n n is us. And why, than X must be less than why in this case, we don't necessarily have to Interesting equations. We do have positive integers, right? So I means K or N whichever variable we use is gonna have to be greater than or equal to one. Right? So 123 all the positive integers We, uh, actually will use that in this question. So, um, let's see what we've got here. P of k. If we'd like, we could be right that also, um, won't necessarily help us very much. But, um, we could it might get us one step closer. Let's take a look Care. So this is the same as if we multiply on both sides by two. We get to to the K I was less than two to the K plus one right or two to the K times to to the one which is the same as to to the K plus one. All right, now coming back down to here, we're kind of stuck at this point. We don't We're trying to find that middle term. Right? So let me leave this up here. What do you use X and Y X is less than it were kind of aiming for this. If you can get both of these parts of equations, then we're kind of golden at that point. Let's use this K is greater than or equal to one. Um, we Yes, there we go. OK, so our middle term right now is is either K right or to K if we wrote it in this this way. Um and so what will help us is using this two K version. All right, so looking at this, this is just the definition for natural numbers. We have assumed that on the question because they told us that, uh, we could add K to both sides if we wanted to. The reason we might do that. So that under left side, we could get to K. Right. Um, and now you can see we have something valuable. We have one plus K that matches here, and we have to k, which also matches right here. All right, so let's write everything else. Everything that we know. We have one plus K, it was less than or equal to two case. Then we have two K is less than two to the K plus one. Okay, so we could say that K plus one must be less than two to the K plus one. Now, this symbol might be throwing me off a little bit. But let's say even if if we prove that this was equal which is possible in this case, alright, If K plus one equals two to the case, it still is less than two to the K plus one. So here we have shown that K plus one is less than two to the K plus one, which is exactly what we were trying to prove, right. That's that's right here were able to prove that.