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Binomial Expansion

  1. Nov 22, 2011 #1
    The coefficient of x in the expansion of [x+(1/ax^2)]^7 is 7/3. Find the possible values of a.

    1. Rewrite (x + 1/(ax^2))^7 = x^(-14) (x^3 + 1/a)^7.
    So, we need to find the coefficient of x^15 from (x^3 + 1/a)^7.

    2. Using the Binomial Theorem, we have
    (x^3 + 1/a)^7 = Σ(k = 0 to 7) C(7, k) (x^3)^(7 - k) (1/a)^k.
    ...................= Σ(k = 0 to 7) C(7, k) x^(21 - 3k) (1/a)^k.

    3. So, we need 21 - 3k = 15 ==> k = 2.

    Thus, we have (1/a)^2 = 7/3
    ==> a = ±√(3/7).

    The problem is, I do not understand the steps. Help please?
     
  2. jcsd
  3. Nov 22, 2011 #2
    First [itex]\frac{1}{x^{2}}[/itex] is taken out as a common factor. Note that this factor is to the power of 7.
     
  4. Nov 22, 2011 #3
    Why do we need to find the coefficient of x^15 from (x^3 + 1/a)^7?
     
  5. Nov 22, 2011 #4

    Ray Vickson

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    How else could you get x^1 when multiplying x^(-14) and x^n?

    RGV
     
  6. Nov 23, 2011 #5
    Oh right.. Thank you! :)
     
  7. Nov 23, 2011 #6

    HallsofIvy

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    I wouldn't have factored out the [itex]x^{14}[/itex]. Each term of [itex](u+ v)^n[/itex] is of the form
    [tex]\begin{pmatrix}n \\ i\end{pmatrix}u^iv^{n-i}[/tex]

    Here, u= x and [itex]a= 1/ax^2= (1/a)x^{-2}[/itex] so that formula would be
    [tex]\begin{pmatrix}7 \\ i\end{pmatrix}a^{i- 7}(x^i)(x^{2i- 14})[/tex]
    and we want the power of x, i+ 2i- 14= 1. That gives 3i= 15 or i= 5.

    It looks to me like you have forgotten the binomial coefficient. I get an integer value for a.
     
  8. Nov 23, 2011 #7
    I don't understand how [tex]\begin{pmatrix}7 \\ i\end{pmatrix}a^{i- 7}[/tex] can come from [tex]u^iv^{n-i}[/tex]

    If it's n-i shouldn't it be a7-i and not ai-7?
     
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