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Binomial Theorem expansion with algebra

  1. Mar 4, 2008 #1
    In the binomial expansion [tex](2k+x)^{n}[/tex], where k is a constant and n is positive integer, the coefficient of x² is equal to the coefficient of x³

    a) Prove that n = 6k + 2
    b) Given also that [tex]k = .\frac{2}{3}[/tex], expand [tex](2k+x)^{n}[/tex] in ascending powers of x up to and including the term in x³, giving each coefficient as an exact fraction in its simplest form.

    my shot at (a)

    expand to get x² and x³:

    [tex]\stackrel{n}{2}(2k)^{n-2}[/tex] + [tex]\stackrel{n}{3}(2k)^{n-3}[/tex]

    subst n = 6k + 2 but i get into more expansion, which i dont think is really going anywhere

    can someone help me out/guide me through (a) please?

    Thankyou
     
  2. jcsd
  3. Mar 4, 2008 #2

    arildno

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    So, do you agree that we must have:
    [tex]\frac{n!}{2!(n-2)!}=\frac{n!}{3!(n-3)!2k}[/tex]

    Clearly, this can be simplified to:
    [tex]\frac{3!2k}{2!}=\frac{(n-2)!}{(n-3)!}[/tex]

    Can you simplify this into your desired result?
     
  4. Mar 4, 2008 #3
    i can get it [tex]\frac{3!2k}{2!}=\frac{(n-2)!}{(n-3)!}[/tex] down to 6k(n-3)=1 but that wont simplify to n = 6k + 2.
     
  5. Mar 4, 2008 #4

    arildno

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    Now, you are muddling!
    Which number is biggest: (n-2)! or (n-3)!
     
  6. Mar 4, 2008 #5
    (n-2)(n-1) /
    (n-3)(n-2)(n-1)

    so (n-2)! is cancels

    so leaves 1/(n-3)

    right?
     
  7. Mar 5, 2008 #6

    arildno

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    No.
    We have: (n-2)!=(n-2)*(n-3)!
     
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