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

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The discussion focuses on deriving a formula for the recurrence relation S_{n} = 2S_{n-1} + (2n-4) with the initial condition S_{1} = 0. Participants clarify that this is a non-homogenous recursive relation and express a desire to find a closed form for S_{n}. The conversation includes a step-by-step breakdown of the derivation process, leading to the conclusion that S_{n} can be expressed as S_{n} = 2^n - 2n. Questions about the origin of the equation and the reasoning behind introducing terms like 2n are also addressed. The thread emphasizes the importance of understanding the types of recurrence relations for proper analysis.
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Homework Statement


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


Homework Equations


S_{1} = 0


The Attempt at a Solution


Somone already showed me a proof.
But it starts:
S_{n} + 2n = 2S_{n-1} +2n-4 + 2n
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}.
S_{n} = 2S_{n-1} + (2n-4)


Homework Equations


S_{1} = 0


The Attempt at a Solution


Somone already showed me a proof.
But it starts:
S_{n} + 2n = 2S_{n-1} +2n-4 + 2n
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 S_{n} in terms of n.

S_{n} = 2S_{n-1} + 2n - 4

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))

Since S_{n-1} + 2(n - 1) = 2(S_{n-2} + 2(n-2)

Then S_{n} + 2n = 2(2(S_{n-2} + 2(n-2)))

Working your way down, you eventually arrive at:

S_{n} + 2n = 2^{n-1}(S_{1} + 2(n-(n-1))) = 2^{n-1}(S_{1} + 2)) = 2^{n-1}(2) = 2^n

S_{n} = 2^{n} /left /left – 2n
 

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