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Binomila theorum-examzz help

  1. Jun 18, 2006 #1
    i don't get this question

    in the expansion of (3/x - x^3)^8 find

    a) the constant term ( is dere an easier way to find this instead of expandin the equation n den finding it)

    b) the term containin x^12
     
  2. jcsd
  3. Jun 18, 2006 #2

    Andrew Mason

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    The expansion of [itex](x + a)^n[/itex] gives terms:

    [tex](x + a)^n = \sum_{k=0}^n \left(\begin{array}{c} n\\k\end{array}\right) x^ka^{n-k}[/tex]

    where the coefficient:

    [tex]\left(\begin{array}{c} n\\k\end{array}\right) = \frac{n!}{(n-k)!k!}[/tex]

    So multiply your expression by -1/x^8 and then do the binomial expansion. Then multiply the binomial terms by -1/x^8 to find the constant term and one containing x^12 (if there is one).

    [tex](3/x - x^3)^8 = (\frac{-1}{x}(x^4-3))^8 = \frac{1}{x^8}\sum_{k=0}^n \left(\begin{array}{c} 8\\k\end{array}\right) x^{4k}(-3)^{8-k}[/tex]

    Can you do the rest from that?

    AM
     
    Last edited: Jun 18, 2006
  4. Jun 18, 2006 #3
    i still kind adont get it..can u plz complete it:blushing:
     
  5. Jun 19, 2006 #4

    HallsofIvy

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    It's just doing the arithmetic now.
    [tex](3/x - x^3)^8 = (\frac{-1}{x}(x^4-3))^8 = \frac{1}{x^8}\sum_{k=0}^n \left(\begin{array}{c} 8\\k\end{array}\right) x^{4k}(-3)^{8-k}[/tex]

    The "kth" term is
    [tex](-3)^{8- k}\left(\begin{array}{c} 8\\k\end{array}\right)x^{-8}x^{4k}= (-3)^{8- k}\left(\begin{array}{c} 8\\k\end{array}\right)x^{4k-8}[/tex]

    a) The constant term has no x term or an x0 term. For what value of k is 4k- 8= 0? What is
    [tex](-3)^{8- k}\left(\begin{array}{c} 8\\k\end{array}\right)[/tex]
    for that k?
    b) For what value of k is 4k- 8= 12? What is
    [tex](-3)^{8- k}\left(\begin{array}{c} 8\\k\end{array}\right)[/tex]
    for that k?
     
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