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

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to derive a formula to determine any value of S_{n}.
    [tex]S_{n} = 2S_{n-1} + (2n-4)[/tex]


    2. Relevant equations
    [tex]S_{1} = 0[/tex]


    3. The attempt at a solution
    Somone 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.
     
  2. jcsd
  3. Nov 11, 2009 #2

    Mentallic

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    Homework Helper

    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.
     
  4. Nov 11, 2009 #3
    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?
     
  5. Nov 12, 2009 #4
    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
    = 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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