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Binominal theorem

  1. Jun 2, 2009 #1
    ive been asked to work out the coefficent of x^33 in the expansion of

    (1/4 2x^3)^17

    i know about (a+b) and ^C_ but im unsure how to get k which is c_k

    i worked off (1/4)^17-k (2x^3)^k

    which give me 17*1/4 =4 1/4 and that is were i went wrong could you tell me a way to work out k


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 2, 2009 #2

    danago

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    What is the only value of k that will produce the x^33 term?
     
  4. Jun 2, 2009 #3
    so your saying some thing as simple as -16 would make the sum work.
     
  5. Jun 2, 2009 #4

    danago

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    Im not sure how you got -16.

    The binomial expansion formula says that:

    [tex](0.25+2x^3)^{17} = \sum \!^{17} C _k (0.25)^{17-k} (2x^3)^k[/tex]

    As k ranges from 0 to 17. For which of those values of k will the x^33 term appear?
     
  6. Jun 2, 2009 #5
    that is what im not sure about i dont know an easy way to find k
     
  7. Jun 2, 2009 #6

    danago

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    Ok, well consider a few different values for k.

    If k=1, then the (x^3)1=x3 term will appear. If k=2, then the (x^3)2=x6 term appears. Can you see now what value of k will make the x^33 term appears?
     
  8. Jun 2, 2009 #7
    ok i see what you mean i did see a pice in the book about it but i was not sure what it ment.thank you i was thinking that both a+b worked together and did my maths around that i see i could look at it as b on its own.
     
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