Deriving a Formula for S_{n} in a Recurrence Relation

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Homework Help Overview

The discussion revolves around deriving a formula for the sequence defined by the recurrence relation S_{n} = 2S_{n-1} + (2n-4), with the initial condition S_{1} = 0. Participants are exploring the nature of this recurrence and the steps necessary to express S_{n} in terms of n.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the introduction of the term 2n in the derivation process and its necessity. There is a discussion about the origin of the recurrence relation and its implications for proving the formula. Some participants are attempting to identify the type of recurrence relation and express S_{n} in a closed form.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications about the nature of the recurrence relation. Some guidance has been offered regarding the classification of the relation, and there is an exploration of the steps involved in deriving the formula.

Contextual Notes

Participants are navigating the complexities of recurrence relations and expressing concerns about the completeness of the information provided. There is an acknowledgment of the potential confusion arising from different types of recurrence relations.

nobahar
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Homework Statement


I have to derive a formula to determine any value of S_{n}.
[tex]S_{n} = 2S_{n-1} + (2n-4)[/tex]


Homework Equations


[tex]S_{1} = 0[/tex]


The Attempt at a Solution


someone already showed me a proof.
But it starts:
[tex]S_{n} + 2n = 2S_{n-1} +2n-4 + 2n[/tex]
Okay, from here you can work it out, but why introduce 2n from the outset? How would I know straight off that this is what to do?
Thanks in advance.
 
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I've barely even touched on such maths, so forgive me if I'm completely off.

Firstly, where did you derive the equation from? What does this equation represent?
If it's just something that you have defined, how can it possibly be proved? By its definition that IS what it's meant to be.
 
nobahar said:

Homework Statement


I have to derive a formula to determine any value of S_{n}.
[tex]S_{n} = 2S_{n-1} + (2n-4)[/tex]


Homework Equations


[tex]S_{1} = 0[/tex]


The Attempt at a Solution


someone already showed me a proof.
But it starts:
[tex]S_{n} + 2n = 2S_{n-1} +2n-4 + 2n[/tex]
Okay, from here you can work it out, but why introduce 2n from the outset? How would I know straight off that this is what to do?
Thanks in advance.

Hi,

This looks to me like a non-homogenous recursive relation. Are you trying to find a 'closed' form for the n'th term of this recurrence relation, or what exactly is your question?
 
sutupidmath said:
Are you trying to find a 'closed' form for the n'th term of this recurrence relation, or what exactly is your question?
Sorry, you'll have to forgive me. Being familiar with where the equation came from I took it for granted. I thought enough information was supplied.

I looked up recurrence relations and I'm pretty confident that this is what I'm looking for, thankyou for that. I'll have to see if the following is a recurrence relation, I noticed in my preliminary search that there are a number of 'types'; which may confuse things for me... I'll look to see if it is the one you mentioned above, given the following information, could you verify the type for me?

Basically, I just want to express [tex]S_{n}[/tex] in terms of n.

[tex]S_{n} = 2S_{n-1} + 2n - 4[/tex]

[tex]S_{n} + 2n = 2S_{n-1} + 2n - 4 + 2n<br /> = 2S_{n-1} + 4n - 4 = 2S_{n-1} + 4(n - 1) = 2(S_{n-1} + 2(n - 1))[/tex]

Since [tex]S_{n-1} + 2(n - 1) = 2(S_{n-2} + 2(n-2)[/tex]

Then [tex]S_{n} + 2n = 2(2(S_{n-2} + 2(n-2)))[/tex]

Working your way down, you eventually arrive at:

[tex]S_{n} + 2n = 2^{n-1}(S_{1} + 2(n-(n-1))) = 2^{n-1}(S_{1} + 2)) = 2^{n-1}(2) = 2^n[/tex]

[tex]S_{n} = 2^{n} /left /left – 2n[/tex]
 

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