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Solve the recurrence relation given by an+1 with 1, n 2 (There Fnun are few approaches here_ One of them is to calculate many first terms and predict general formul...

Question

Solve the recurrence relation given by an+1 with 1, n 2 (There Fnun are few approaches here_ One of them is to calculate many first terms and predict general formula and then prove the formula by Induction. Another approach is to make transformation, bn S0 that the recurrence becomes linear one |

Solve the recurrence relation given by an+1 with 1, n 2 (There Fnun are few approaches here_ One of them is to calculate many first terms and predict general formula and then prove the formula by Induction. Another approach is to make transformation, bn S0 that the recurrence becomes linear one |



Answers

Find the solution of the recurrence relation $a_{n}=$ $2 a_{n-1}+3 \cdot 2^{n} .$

So here we have the t r Function T is gonna be equal to end at times t squared if I did it on over to and initial condition T one is equal to six. So first put him in a substitute and equals two to the power of K. And so this becomes t of to the power of K is gonna be equal to two to the power of K Times T is squared to the power of K over too. Now, which, of course, is gonna be equal to two out of K times T square Heard it too. K minus one. Um and they're gonna make another substitution here. Where were you? Uh, a Okay, equal blob thio t two to the power of K. And when we do that, we get that this is equal to Bob, too. Two to the power of K Times T squared to K minus one. Yeah, we can use some of our log rules. We get love two times two to the K plus Ah, two times allowed. Two of key to the Caymans. One square. Okay. And this is gonna be equal to okay. Plus two, uh, two tee times two to the cameras, one power. Um and then this is equal to K plus two. Ah said K minus one. Yes, that's gonna be a lot easier to work with. What we've made The substitution, sze and for simplicity, we can really write this as, uh, a n a k would be a n is equal to end plus 2 a.m. I s water. I shouldn't be in the same form that we're used to you. I wanted to translate our initial conditions as well, so we would get so gonna rewrite. To get a N is equal to to a and minus one plus on. And then we would have a knot is equal to log to t to the two to the power of zero, right, which is gonna be just one two to the power becomes t one, which we know. Um, from what we're given is that that is really just steaks. That's what we want to find the roots characteristic equation here on when we do that, we get that our homogeneous portion is gonna be equal to. And so I would say our square is equal to two. Are is equal to two when we set an equal to our own and my 21 s of the Alfa Times two to the power of n And then we want to move on to the particular solution. Uh, in this case, our function is just equal to end. Uh, which is the same as one? Well, to find that by one of the power of an and one is not a route. So we'll take that form of our particular solution but should be equal to P one times and plus p not times one of the part of end, uh, which is just p one times and plus p not. And then now we need Thio Saul for this And make sure it satisfies ari current relation here. And so we will rewrite it. Um, we have a n equals two times a and minus one plus n, but put in our particular solution p one times n plus p not and for a m. We'll put it in for a M s. One is well, so we get to times p one times and minus one plus peanut plus n. We can extend that. So it's gonna be two p one times and the minus two p one plus two p not plus n. I think in some tracks from the left hands high and the rice we would get zero is equal to p one times end minus two p one plus p not plus. And so I get up the ends and the not ends we get. Zero is equal to P one plus one times n plus negative two p one plus peanut, and these coefficients must be equal to zero. So we'll just set them equals zero and solve. And when we do that, we have that p one plus one is equal zero and negative. Two p one plus peanut is equal zero. So the first one's pretty easy, right? So here we get that p one is equal to negative one. We put negative one in here and we end up with two. Plus, peanut is equal to zero. Assis means that peanut is equal to negative too. So we'll put that been to our particular solution, right? Sue? Our particular solution was looking like p one times on the plus p not and so we know this will be negative and minus two. The entire solution then, is going to be our home on genius portioned plus our particular portion. So that is going to be Alfa Times two to the power and minus minus and minus two. So we'll use our initial conditions to evaluate it at an equal zero and fine Alfa. And so at n equals zero we get Alfa minus two, which was equal to a knot was going to be lot two of six. Okay, so that means Alfa is equal to God. Thio six plus two eso es and is going to be We'll have to six plus two times two to the power and minus end minus two. Okay, but now we need to put this back in two terms of tea. Pray and so a N is really equal to log thio t to to the end. But we had called it okay previously, so that is finding their way. Eso we'll say log two t uh, two to the power of K. It's gonna be equal to love to six times to power and plus two to the n plus one. So I just before I had both those multiplied by that to the part of N. And now I have just multiply it out those components a minus and minus two. OK, we could take the exponential of both sides. So we get T T k is equal to two to the power of Glod to six times to the end, plus two plus one minus and minus two. I suppose we shouldn't use thes ks. Okay. And when we do that, this is going to be equal to six to the power of two out of K times two to the power of twos of Howard K plus one times two to the power of negative K times two to the power of negative too. And over the algebra this is equal to, um, can put thes below rights would be six the tower of two to the power of K times two part of two to power K plus one over two to the power of K a times two squared, and that is going to be equal to six to the powder toot out of K times four to the power of two to the power of K over four times two to the power of canceling up here for about a two to the power of two. Uh, so definitely messy. For the moment, we have t two to the power of K being equal to six to the power of two to the part of K times four to the power of two to the power of K over four times two to the power of care. That's enough. We let end be equal to two to the power of K. So that was our first substitution we made. We end up with T power tea. Then it is equal. Thio, what we have up here. Ah, we get six and times for the power of n over four times n which will simplify as six end times for the power N minus one over and

So we need to show that this series and this uh recursive series are the same first. We have that disregards the series. Its first value is one. Therefore S. R. one will also vehicle to two times 1 squared -1. She's just going to be one. This first part is proven. Now we need to prove that for some S. F. K. In the top equation we're gonna have to Kane scored minus K. We also need to find S. FK -1. Should be two times A -1 Squared minus k -1 in parentheses. So um that calculated what's going to become two times K squared minus four K plus two minus k plus one. That's gonna leave us me to case where uh okay Squirt -5 K. Last three. So um having that now we need to prove that it's a hey Physical to S. F.K. Last four k minus three. So let's see if that's true. It's okay. Was two K squared minus K and it's ok minus or it will be two K Squared -5 K. Plus three. And then there is a plus four k minus three. So plus year 93 cancels out and the four K. And a negative phi k lives a negative kid. So we see that they are the same. So therefore we have proven that the recursive equation is just this one

We wish to show that this sequence here satisfies this recurrent recurrence relation. Okay, so there are two things we have to check. You have to make sure that the first term matches. And then we had to check that this relationship holds. So let's first check this when using the formula S one is going to be one times one plus 1/2 which is 2/2 and that's equal to what. So that's good. Now we put in this formula for sn into this recurrence relation. So s. n. on the left side, that's going to be 10 times And plus 1/2. And then we need to check, is that equal to the left hand side? Which is S uh minus one plus N. So that's what we And -1 times and minus one plus one, which is 10 over to. And now we add on. So let's simplify the right hand side. So well put it over common denominator that's going to be and -1 times and plus two N over to. Okay, so now let's actually distribute that. That's going to be unscrew aired minus and Plus two and which simple advice to just press on over to. And then you can see that if we actually who Factor out the end that's 10 times and plus one over to which is exactly the same as the left hand side.

We wish to show that this sequence here satisfies this recurrence relations. So to do that we need to first show that the first term is the same. So S one using the formula Is equal to two times 1 sq minus one. So this is 2 -1, which is equal to one. So that is good. Now what we would do is we would use this formula and put it into this recurrence relation to see if the left side is equal to the right side. So we get S and on the left side, so that would be too and squared minus. And and then we need to see is that equal to the right hand side? So s and minus one should be two times and minus one squared minus and minus what? And then plus for an minus. So now all we have to do is simplify the right hand side and see if we end up with an expression that's equivalent to the left hand side. So let's actually expand this out. This is going to be two times and this will be a nice squared minus two N plus one. This is minus an plus one Plus four and -3. Case of this is going to be too and squared minus four N plus two and then minus N plus one plus four and minus three. So now let's group the like charmers together. We have to and squared and then we have negative for end, positive foreign and uh minus end. So that's minus. And and then her constant would be two plus one minus three, which is zero. So now we can say that uh this recurrence relation holds because the left hand side is exactly the same as the right hand side.


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