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Math Help - Binomial expansion question

  1. #1
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    Binomial expansion question

    Hi everybody,
    I am having some trouble with these questions and I'm not really sure where to start.


    Find the term independant of x in the expansion of:

    <br />
                         (\frac{1}{3x} - \frac{3x^2}{2})^9

    Find the value of n for which the coefficients of x, x^2 and x^3 in the expansion of (1 + x)^n form three consecutive terms of an arithmetic series.

    Thanks all.
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  2. #2
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    Quote Originally Posted by mathsnerd1 View Post
    Hi everybody,
    I am having some trouble with these questions and I'm not really sure where to start.


    Find the term independant of x in the expansion of:

    <br />
                         (\frac{1}{3x} - \frac{3x^2}{2})^9
    The general term of (a+ b)^n is \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n-i}.
    Here, a= \frac{1}{3x}, b= \frac{3x^2}{2} and n= 9 so that is \begin{pmatrix}9 \\ i\end{pmatrix}\frac{1}{3^ix^i}\frac{3^{9-i}x^{2(9-i)}{2^{9-i}}

    Now, for what value of i is \frac{x^i}{x^{2(9-i)}}= 1?

    Find the value of n for which the coefficients of x, x^2 and x^3 in the expansion of (1 + x)^n form three consecutive terms of an arithmetic series.
    Those coefficents are, of course, \begin{pmatrix}n \\ i\end{pmatrix}= n, \begin{pmatrix}n \\ 2\end{pmatrix}= \frac{n!}{2(n-1)!} which must be equal to n+r, and \begin{pmatrix}n \\ 3\end{pmatrix}= \frac{n!}{3!(n-3)!} which must be equal to n+ 2r, for some r, in order to form an arithmetic series. that gives you two equations to solve for n and r.

    Thanks all.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    The general term of (a+ b)^n is \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n-i}.
    Here, a= \frac{1}{3x}, b= \frac{3x^2}{2} and n= 9 so that is \begin{pmatrix}9 \\ i\end{pmatrix}\frac{1}{3^ix^i}\frac{3^{9-i}x^{2(9-i)}{2^{9-i}}

    Now, for what value of i is \frac{x^i}{x^{2(9-i)}}= 1?

    Where did i come from?

    And how did you get \frac{x^i}{x^{2(9-i)}}= 1?
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