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Binomial Theorem question

  1. Mar 28, 2005 #1
    Hi, I'm having some problems with this question -


    Find the term independent of x in

    (x^2 - 2/x)^6


    I know the answer is something like 6Csomething 2^something, but I'm not sure how to get that. So far I've only really done simple things like (x+y)^n where y is an integer and not another x term. Help would be appreciated.
     
  2. jcsd
  3. Mar 28, 2005 #2

    Zurtex

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    Well the long way to do is expand the full thing. But there are easier ways to do it, for example rewrite it as:

    [tex]\left(x^2 - 2x^{-1}\right)^6[/tex]

    Expand that without looking at the coefficients (and I inculde the sign of the term when I say coefficient) and you get the general expansion:

    [tex]a\left(x^2\right)^6 + b\left(x^2\right)^5 x^{-1} + \ldots[/tex]

    Simplifying:

    [tex]ax^{12} + bx^{9} + \ldots[/tex]

    Carry on like that until you get a x0 term and work out what the coefficient is.
     
  4. Mar 28, 2005 #3

    dextercioby

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    HINT:

    [tex] (a-b)^{n}=\sum_{k=0}^{n} C_{n}^{k} (-1)^{k}b^{k}a^{n-k} [/tex]

    Answer:k=4...

    Daniel.
     
  5. Mar 28, 2005 #4
    simplify the equation :
    f(x) = h(x).g(x)

    [tex](\frac{2}{x})^6 ({1- \frac{x^{3}}{2})^6[/tex]

    Use the binomial expansion for g(x) and simplify the solution
     
  6. Mar 28, 2005 #5

    Zurtex

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    That's a simpler equation :confused: ?
     
  7. Mar 28, 2005 #6

    dextercioby

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    My formula delivers the result in 2 lines,on of which is

    [tex] -k+12-2k=0 [/tex]

    Then computing [itex] C_{6}^{4} [/itex] is elementary.

    Daniel.
     
  8. Mar 28, 2005 #7

    xanthym

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    .

    And for those who would rather not compute [tex] \mathbb{C}_{4}^{6}} [/tex], there's always Pascal's Triangle:

    Code (Text):


    [B]
                           1
                         1   1
                       1   2   1
                    1    3   3    1
                 1    4    6    4   1
               1   5    10   10   5   1
             [COLOR=Red]1   6   15   20   15   6   1[/COLOR]
    [/B]
     

    ~~
     
    Last edited: Mar 28, 2005
  9. Mar 28, 2005 #8

    dextercioby

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    Rather than drawing the Pascal triangle,don't u think it's easier to multiply 6 by 5 and divide the result by 2...?

    Daniel.
     
  10. Mar 28, 2005 #9

    xanthym

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    For n<10, Pascal's Triangle is easily derived in 10 sec and serves the needs of:
    1) Those who would take longer than 10 sec to remember that [tex] \mathbb{C}_{4}^{6} = (6!)/[(4!)(2!)] [/tex] ; or
    2) Those who want ALL the coefficients in 10 sec; or
    3) Those who compulsively doodle.


    ~~
     
  11. Mar 28, 2005 #10

    dextercioby

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    I could multiply & divide natural #-s without using a paper.Imagining Pascal's triangle & doing the addition and having the overview is harder,for me,at least...:wink:

    Daniel.
     
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