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Binomial expansion

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the expansion (ax2 + bx + c)n = ∑(r=0 to r=2n) Ar xr------------------(1) , where Ar is real ∀ 0 ≤ r ≤ 2n
    Replacing x by c/(ax) and using the property ∑(r=0 to r=2n) Tr = ∑(r=0 to r=2n) T2n-r ,
    we get (ax2 + bx + c)n = ∑(r=0 to r=2n) Br xr ----------------------(2).
    Hence we get a relation between Ar and Br by comparing like power of xr
    0 ≤ r ≤ 2n from both equations.

    Let (2x2 + 3x + 4)10 = ∑(r=0 to r=2n) Ar xr then the value of
    A6 /A14 is

    2. Relevant equations

    3. The attempt at a solution
    I tried doing what the paragraph says--
    I substituted (2/x) for x (c/ax = 2/x).
    I got (8/x^2 + 6/x + 4)10
    But I don't want a relation between A and B
    I want the ratio of the coefficients of the terms with x^6 and x^14.
    I tried doing it using binomial expansion by making cases for when the powers of (4+3x) and (2x^2) add up to 6 and 14.
    For 6, I got 4 such cases. For 14, I got 8. However, solving all the Cs and stuff is very time consuming and doesn't seem the right way to go.
  2. jcsd
  3. Nov 10, 2014 #2


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    Homework Helper

    The important thing to recognize is that your ##B_r## should be equal to your ##A_r## since they are expanding the same polynomial.
    By introducing the substitution, you can find a way to relate ##B_{2n-r}## to ##A_r## which works for r=6.
  4. Nov 10, 2014 #3


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    Homework Helper

    Keep it in the sum notation to see the relationships.
    ##\sum_{r=0}^{2n} A_r x^r ##
    ##\sum_{r=0}^{2n} B_r \left(\frac{c}{ax}\right)^r=\sum_{r=0}^{2n} B_r \left(\frac{c}{ax}\right)^{2n-r}##
    Do some algebra to equate the powers of r, and see what comes out.
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