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THE RECURRENCE RElATion: consjPer 30 10 On For na 0,1,2,3, 0n+2 ntl W #ere Qp= | 4 = 2 COMPL ETE THE TKBLE 1 2/2/ 4 2 DETERMiNE A CLOSED FORMULA AND C HEck IT AGLE...

Question

THE RECURRENCE RElATion: consjPer 30 10 On For na 0,1,2,3, 0n+2 ntl W #ere Qp= | 4 = 2 COMPL ETE THE TKBLE 1 2/2/ 4 2 DETERMiNE A CLOSED FORMULA AND C HEck IT AGLEES WiTh Your FOR Qn TA BLE Your CLOSED FORMULA For an uSE To DE Ter Mine AND (0

THE RECURRENCE RElATion: consjPer 30 10 On For na 0,1,2,3, 0n+2 ntl W #ere Qp= | 4 = 2 COMPL ETE THE TKBLE 1 2/2/ 4 2 DETERMiNE A CLOSED FORMULA AND C HEck IT AGLEES WiTh Your FOR Qn TA BLE Your CLOSED FORMULA For an uSE To DE Ter Mine AND (0



Answers

Find the solution of the recurrence relation $a_{n}=$ $4 a_{n-1}-3 a_{n-2}+2^{n}+n+3$ with $a_{0}=1$ and $a_{1}=4 .$

You're given a recurrence relation from your ass Defying to the solution of this relation their occurrence relation is a N equals 2 a.m. minus one plus three times two to the end. This is a linear, non homogeneous recurrence Relation the associate ID When your ma Jenness recurrence relation is a N equals 2 a.m. minus one And here we have the characteristic equation is ar minus two equals zero said the characteristic route. It's simply r equals two we have that the general solution to this homogeneous equation has the form Alfa Times two to the end where Alfa is some constant, we have the FN. The function of end is three times two to the end three's a polynomial of degree zero and we have it too, is a characteristic group of multiplicity one and so it follows. That's the general form. Very particular solution. It's going to be end times p zero times two to the end and to find the value the coefficients we put particular solution back into the non homogeneous equation. So we have p zero end times two to the end is equal to two times p zero times n minus one times two to the N minus one plus three times two to the end. If we divide those sides by to to the n minus one, we get two he not and is equal to to p zero and minus two p zero plus six in writing this on one side we have zero is equal to two p zero ends, cancel out and we have negative two p zero plus six. And so it follows that p zero is going to be equal to three so that the particular solution is three n times two to the end and therefore we have the general solution for this non homogeneous equation is the general solutions. The modernise equation must particular solution and this is equal to Alfa Times two to the end, plus three end times two to the end.

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In part a were given a recurrence. Relation in grass to find all solutions recurrence Relation is a n equals 2 a.m. minus one plus two n squared. This is a linear, non homogeneous recurrence relation and so the associate ID When you're homogeneous recurrence relation is a N equals to a and minus one. We have that The characteristic equation is ar minus two equals zero so that the characteristic route is to and therefore we have that the General form 40 solution to the associate ID linear homogeneous recurrence relation is Alfa Times two to the end where alphas some constant. And we have that in the non homogeneous equation FN is equal to and squared we have This is the same as and squared times one to the end One is not a characteristic roots we have the end squared is a pulling on your degree too. So it follows that the general form for a solution just particular is going to be p two and squared plus p one n plus p zero times one to the end which is just one and we have that all solutions are given by some of the particular solution and the solution to the associate ID homogeneous linear recurrence relation, which is going to be my mistake. They should be plus two and squared ref Event is not in squares to m squared, so to determine the coefficients on the particular solution plug it back into are not imagines equation. So we have it. P two and square plus p one end plus P zero is equal to two times, and this is going to be Pete two times in minus one squared plus P one times in minus one plus P zero and all this plus two one squared we can write. This felt terms on one side, so we have P to end squared plus p one end plus P zero is equal to two p two times and squared and then minus two n plus one plus p one times and minus one plus P zero plus to end squared from which we obtain p two and squared plus p one n plus p zero is equal to two p two and squared. But this is to be too and squared, plus two and squared and then for end. We have a coefficient negative four p two and we have minus two p one times. And and for the constant term, we have two p two minus P one minus two p one I mean and plus two p zero. And so we have that. Zero is equal to p two plus two and squared plus says negative for Pete to minus three p one times in and then plus two p two minus two P one and plus P zero. And it follows that the coefficients of the polynomial are all zero. So we have that. P two is equal to negative two. We have that p one he's going to be able to or p two over negative three. So negative 4/3 P two, which is equal to negative. 4/3 of negative, too, which is 8/3. And Pete, too, is equal to 1/2 of this is negative. Not too, though I'm selling p zero. My mistake. P zero is equal to negative. Two p two plus two p one, which we have is equal to negative two times Negative two, which is four plus two times 1/3 which is 16 3rd This is gonna be 12 with 16 is 28 3rd I made mistake. That should have been a plus two p one here to this becomes lyrical people's two plus negative for P two plus three p one or in this case, just plus P one, and that we have a two p two minus two p one still equal to two p zero house two p zero. So we have p two is negative. Two p one is four p two, which is negative. Eight and P zero is negative. Two p two plus two p one is going to be negative. Two times negative two is four plus two times negative. Eight is negative 16 four minus 16 which is negative. 12. And so we have. The particular solution is negative to end squared plus eight n minus 12. And so we have that the general solution is the particular solution. Plus okay, solution to the associate ID Ma Jenness equation. So this is going to be Alfa Times two to the end, minus two and squared. We should be in negative eight and it's minus joined square to minus eight in minus 12. This is our solution for part A in part B for us to find a solution of their occurrence relation from part A. With the initial condition, that's a one is equal to four. It's using the general form for a solution we have. The four is equal to health. UH times two minus two, minus eight minus 12. This means that to Alfa he's going to equal to six. Plus eight is 14 plus 12 26 where the Alfa is equal to 13. And so it follows that the solution is 13 times two to the end, minus to end squared minus 8 10 minus 12.


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