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Math Help - recurrance relation

  1. #1
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    May 2009
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    recurrance relation

    I'm working on this R.R.:

    a_n = a_{n-1} + n(n+1)
    a_0 = 3

    when I get to this part of the soln, I am unsure how to get from
    a_n = a_{n-1} + n(n+1) = a_{n-1}+n^2+n
    .
    .
    .
    = -2(3^{n-1} + 3^{n-2}+ 3^{n-3} + ... + 3^1 + 3^0)
     = -2(\frac{3^n+1}{3-1}) = 1- 3^n

    specifically, how do you derive the recurrence? I don't understand where the denominator comes from
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  2. #2
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    Quote Originally Posted by nowrd2xpln View Post
    I'm working on this R.R.:

    a_n = a_{n-1} + n(n+1)
    a_0 = 3

    when I get to this part of the soln, I am unsure how to get from
    a_n = a_{n-1} + n(n+1) = a_{n-1}+n^2+n
    .
    .
    .
    = -2(3^{n-1} + 3^{n-2}+ 3^{n-3} + ... + 3^1 + 3^0)
     = -2(\frac{3^n+1}{3-1}) = 1- 3^n

    specifically, how do you derive the recurrence? I don't understand where the denominator comes from
    = -2(3^{n-1} + 3^{n-2}+ 3^{n-3} + ... + 3^1 + 3^0)=-2\Sigma_{k=0}^{k=n-1}3^k which is a finite geometric series so it =-2\frac{1-3^n}{1-3}=1-3^n
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