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Prove the following statement using mathematical induction. Donot derive it from Theorem 5.2.1 or Theorem 5.2.2.For every integer n ≥ 1, 1 + 6 + 11 + 16 + ...

Question

Prove the following statement using mathematical induction. Donot derive it from Theorem 5.2.1 or Theorem 5.2.2.For every integer n ≥ 1, 1 + 6 + 11 + 16 + + (5n − 4) = n(5n − 3)2.Proof (by mathematical induction):Let P(n) be the equation1 + 6 + 11 + 16 + + (5n − 4) = n(5n − 3)2.We will show that P(n) is true for everyinteger n ≥ 1.Show that P(1) is true: Select P(1) from the choices below.1 + (5 · 1 − 4) =1 · (5 · 1 − 3)1 = 1 · (5 · 1 − 3)2 P(1) = 5 · 1 − 4P(1)

Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. For every integer n ≥ 1, 1 + 6 + 11 + 16 + + (5n − 4) = n(5n − 3) 2 . Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + + (5n − 4) = n(5n − 3) 2 . We will show that P(n) is true for every integer n ≥ 1. Show that P(1) is true : Select P(1) from the choices below. 1 + (5 · 1 − 4) = 1 · (5 · 1 − 3) 1 = 1 · (5 · 1 − 3) 2 P(1) = 5 · 1 − 4 P(1) = 1 · (5 · 1 − 3) 2 The selected statement is true because both sides of the equation equal . Show that for each integer k ≥ 1, if P(k) is true, then P(k + 1) is true : Let k be any integer with k ≥ 1, and suppose that P(k) is true. The left-hand side of P(k) is ---Select--- 5k − 4 1 + (5k − 4) 1 + 6 + 11 + 16 + ⋯ + (5k − 4) , and the right-hand side of P(k) is . [The inductive hypothesis states that the two sides of P(k) are equal.] We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + ⋯ + (5(k + 1) − 4) = . After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes ---Select--- k(5k − 3)/2 ((k − 1)(5k − 3))/2 ((k + 1)(5k − 3))/2 ((k − 1)(5(k − 1) − 3))/2 + (5(k + 1) − 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal . Hence P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]



Answers

Use mathematical induction in Exercises $3-17$ to prove summation formulae. Be sure to identify where you use the inductive hypothesis.
Let $P(n)$ be the statement that $1^{2}+2^{2}+\cdots+n^{2}=n(n+$ 1$)(2 n+1) / 6$ for the positive integer $n .$
a) What is the statement $P(1) ?$
b) Show that $P(1)$ is true, completing the basis step of a proof that $P(n)$ is true for all positive integers $n .$
c) What is the inductive hypothesis of a proof that $P(n)$ is true for all positive integers $n ?$
d) What do you need to prove in the inductive step of a proof that $P(n)$ is true for all positive integers $n ?$
e) Complete the inductive step of a proof that $P(n)$ is true for all positive integers $n$ , identifying where you use the inductive hypothesis.
f) Explain why these steps show that this formula is true whenever $n$ is a positive integer.

So we're going to prove that bond in your bedroom By Mark Maske. Lynn's Ocean A Wiggins verified the n equals one. So this lets on size. It's just going to be able to speak to her of one is just a plus B the right hand size. Every first term is going to be one of zero, which is one times AIDS, but one beats his ear. Plus second, Sam is going to be +111 80 This one who's able speak yes, n equals one case. Next we're going to replace. And okay, so we're assuming this is true. That's right. Uh, okay. Janessa be, uh, So I'm assuming this is true. And then we're going to show what we needs. Proof? A plus B power caper equals, but some lipstick A plus one from our equals zero. Okay. Plus one a que plus one minus. Uh, the, uh This is what we're trying to prove. Yes, that is Poppy. No, it's this part. See, we're going to do is most played both sides of the n equals K case by April, so let's downsize is easy. We're going to have it. Must be okay. Was bought by a plus B, which clearly is a plus B two, the cape Lis Wiehl Just for my experience. Reckon size is difficult. We're going to get a plus B times, some from Arco zero. Okay. Okay. Uh, okay. Came on. So be to be off. Andi recommends distributing a B, and I'll show you what this means. So first time at this is okay of zero most played by eight. Okay. Oh, he's just be 20 Service is up first. Um, when was playing it by a plus B. So we have most part by a gives us a get this one. We used to have Kim's ever A to the K minus, huh? Most. Why would they? Oh, except our reserve get rid about status of Eisuke. Most blood Monday. So that is the first term of this some most part by April. Speak. This is the most public A This is B Steve out from a second, Sam, that's going to be the okay one we're gonna get. So this would previously be came on this one. Now it's just going to be okay. There's gonna speeds for one. That's the second son was supplied by a on the second son was Plan B is going to be eight came on this one be squares. Right. So this is going to continue, But I can see here a common factor, Like in a place. My two cents here with, uh, this, like so. And as you can see, the cave zero plus que of one from one of the previous questions is equal to okay, plus one for the one Because the last question asked us to prove that's, uh, plus, and of our pulse one, he was endless. One plus one. Um, we now see why? Because this is useful it for proving induction. So I just make some space. I can the place this sum here with Kay plus one of one. Hey, it's okay, okay. And you can see that this is going to continue because we've got the first we'll hear the next bit is going to be que Ever two aids became on this one piece was we could do the same to them to get que plus 1/2 A to the K one. This one piece worth. And this looks suspiciously like the 1st 3 terms of our Capel's one case. I keep this one side. Okay? And look, it's our K plus one case, I guess K plus one of ah A to the K plus one minus up. Beats me up. First up is going to be okay. Plus one zero was fought by a need to take a plus. One second some is going to be okay. +11 over A It's okay, baby, but some is going to be okay. Plus one, two, uh, A to the K minus one be squares. And so until he hits. And, um so you see, it's all cleans up nicely to give us our complete solution.

Here question is asking about the my people thinks about the binomial expansion off a plus B to the power and which ISS and C zero multiplied by eight to depart and the NC one air to depart and minus one multiplied by B plus like this, it will have NC and my people but here to be part of your beauty power. And so the first part of this question is asking the value for an Is it questo one. So when we will put an ethic was toe that a place beauty part and will be a plus B only. And then the second part of this question is asking that replaced and with Kay and write this statement that is assumed to replace and with K plus one and write the statement that must be pulled. Okay, So when you will use the mathematical induction, then first they will put an Is it quest. Okay, which we will assume as true. And then we will put a plus one which need to be approved. So for any is equals to pay that tongue, Or you can say the biometrics pension. Maybe like this Casey Ciro multiplied by a to the Power K plus casing one My Deep, led by a J D Power game minus one B less Casey. Okay, multiplied by B to the power play Now Similarly, for an equals toe K plus one, the town will be like this K plus one. See Zito A to the power K plus one plus a a plus one c +18 repower que and be similarly K plus one c k plus one b to the ballot, shapeless one. So these two are the term. Which question is asked? Not the third. Part of the question is saying that my people I bought the sides of the statement assumed to be true. It means energy quest. Okay, wit a plus B. So when we will multiply a plus B with a plus B to REpower K, which is nothing but a plus B to the Power K plus one. So the question is saying exponents on the left on the right, distribute a plus B respectively. It means multiply this time with a plus B and finally Pruitt with this so we can multiply this with a plus B and we will get this one this is the card, but no, the fourth part of depression is asking the leg like terms on the right. At this point, you should have a plus. B Republic K plus one is equals to K C zero multiplied by a to the Power K plus one plus K C zero plus Casey one multiplied but a to the power k b. So similarly, we will have the towns expanded, or here, which is we can get through the multiplication off a plus B into a plus B to the power. Okay, now, the fifth part will against it. A six part similarly asking about the expansion off this binomial theorem and some terms which are anti are plus N c r plus one like this. It was toe emplacement unpleasant, which we have proved earlier. And the question is again asking that case is you is a quest. Okay. Plus one c zero or not, or a city or region close to cape. Listen, see cables on Is it west or not? Yes, it is. It was too, because KC zero is one only and a +10 is also one only. And similarly, KCET Gaisal toe one and shapeless on CK politicize one. So this is the total question which we have discussed over here and complete answer of this question.

In this question, we are asked to prove by induction the statement that said the some off interchanges Q from one to end equals this term in time plus one over two square. So we will call this statement pee off in for in as some posted deviant ages. My using induction, we have to step two follows Firth. The basic step is the first case that this that men holds true and in this case is when in equal to one, that's not always the case that it has to start with one. Just that in this case, one is the basic step. Ah, other case may start from 25 11 depending on questions. So don't don't ah, stuck on this and equal to one always. Uh, okay, so P off one, we just substitute in s one. Right. So on this side, there will be some off just 12 And on this side B, we have this number is two one square. So is one right? That means P one said one equals to one. So it's true. Then we're clear. This best extent. Next the inductive step. We assume that if the state men is true for For some in that we won so is greater or equal to one some positive integer We want to show that and plus one is true as well. So this hasn't happened yet. We have to show it and how to show it. We can use peon ass the starting point because it is like and assume statement, so we have it as facts as information. Now consider the sum from 1 to 10 plus one Q light this first end term, you can see that we can use inductive hypotheses, right, Because is in is up to in so p of inward that mean we can change it into this term and the last term in status saying now we we rearranged this Be pool in plus one square out. This will be in squire over four plus in past one, and we just add this fraction to the hotel and you can see that the Theobald term can be factor into in plus two square. This is in plus two square. So that means all in all the whole term become this last formula which, luckily, is that the state men four p folk pee in plus one. Right. The lock left hand side is there some up to in 1st 1 and the right hand side is in plus one in plus two. So now we have shown that you're seeing you seeing what we assume. The next statement, the N plus one is also true. When we clear this two step, then we can conclude by men induction that that the statement is true for for any in and the post a TV and take you in. And that is how you do it. Hope this is helpful. Thank you.

High in this question. We are asked to prove using induction the formula, the inequality regarding it's some and this formula. So we one end to start with to like, at least two. And here we go. We let this whole state men BP au vin the basic step and equal to two we have on the left hand side of some off one and 1/4 on the right hands. I in the formal Abu have to minus Uh huh. And this will become so five over full, less than tree over too. Wishes correct, Right? So the basic step is true. Next for the inactives that we supposed us that men is true for some in and we want to show that the in plus one is true s bill. And one way to do this is to start with P au vin here. So we have this inequality and IBU ad this term on both sides so that the some on the left become the sum we want in the in. Plus one statement What's left is to ling like is to show that this right hand side is less than the farm lobby. One right and How do I dump from there to there? Well, we let's look at this. I claim that that this inequality is true. Why? Because let's see, this term 10 n plus one square is less than one away in plus one, right? Because in this positive integer and one over in plus one is if I multiply in on, like, above and below, so it doesn't change anything. I can break this into this to turn one over in minus 10 n plus one. And this this whole lie off a human man, the a bold claim. True, because we can just move one over in to the left side, off the inequality and that that's make us this inequality. And so we use this Indy in the previous that that'd be dumb from that to there. And oh, in all this showed up the statement when off p and plus one is true, right? And so we have proven the inactive state as well. And so with this two step, we have proven the Steadman fall. Any positive integer in greater than one on That is it. Thank you


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