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Finding value independent of x in a binomial expansion?

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    A = 420
    k = 2

    These were worked out from the earlier part (a)

    What is the method of calculating the independent term ? Someone plz help

  2. Offline

    Find for r=5 (this I did by recognition and some thought...dont really think there is a 'method')

    edit: So
    15C5 k^5 = 3003*32= 96096 I think...
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    ^{15}C_5(x^{10})(\frac{2}{x^2})^  5 = 96096

    I agree with the above post too, just find an r value so that the x terms cancel.
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    what is the r?
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    (x+\frac{k}{x^2})^{15} = \sum_{r=0}^{15} {15 \choose r}x^{15-r}(\frac{k}{x^2})^r
    that is, the fomrula for a binomial expansion, putting n=15

    So the general term is {15\choose r}x^{15-r}(\frac{k}{x^2})^r = {15\choose r}x^{15-r}(kx^{-2})^r

    Can you find a value of r such that the power of x is zero? (Get the x indices together and solve the equation, or just do it by inspection). Evaluate the general term for this value of r.
  6. Offline

    nCr. Number of ways of choosing r objects from n items.


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Updated: May 15, 2007
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