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How do i factor that induction expression?

  1. Oct 14, 2009 #1
    Hi,

    i need to prove that [tex]\sum^n_{k=1}\frac{k^2}{2^k} = 6-~\frac{n^2+4n+6}{2^n}~~(*)[/tex]

    for n=1 the statement holds as [tex]\sum\frac{1}{2} = 6 -\frac{11}{2}[/tex]

    [tex] \sum_{k=1}^{n+1} \frac{k^2}{2^k} = 6-~\frac{n^2+4n+6}{2^n} + \frac{n+1}{2^{n+1}}[/tex]

    In order to get a common denominator multiply (*) by 2

    [tex]\frac{2n^2 + 8n +12 + n^2 +2n +1}{2^n} \longrightarrow \frac{3n^2 + 10n +13}{2^n}[/tex]

    But how do i factor that expression ?
     
  2. jcsd
  3. Oct 14, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: induction

    You have a mistake here. Your sum adds numbers of the form [itex]k^2/2^n[/itex] so the term added should be
    [tex]\frac{(n+1)^2}{2^n}[/tex]

     
  4. Oct 14, 2009 #3
    Re: induction

    okay,

    So now we have [tex]\frac{2n^2+6n+7}{2^n}[/tex]

    But shouldn't the new expression be of the form [tex]\frac{\text{polynomial}}{2^{n+1}}[/tex]?
     
  5. Oct 14, 2009 #4

    lanedance

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    Homework Helper

    Re: induction

    i haven't checked your answer, but to get in the form you want, you could you try multiplying by 2/2 then rearranging the polynomial in terms of (n+1)
     
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