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Math Help - Binomial Expansion?

  1. #1
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    Binomial Expansion?

    Hi, I am having trouble with the following queston:

    The constant in the expansions of (Equation 1) and (2) are equal, and k and p both greater than zero. Express k in terms of p.

    Equation 1
    ( kx^2 + 6/x^2 ) ^4

    Equation 2
    ( kx^3 + p/x^3 ) ^6


    It at the end of a chapter on sequences, series and binomial expansion, I don't really understand what the question is asking me to do... I presume I should expand both equations, and that the final answer will be k=...

    Why the two equations though? Are the equal to each other?
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  2. #2
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    Hello, oupib!

    The constant in the expansions of expression (1) and (2) are equal, and k, p \,>\,9

    Express k in terms of p.

    Expression (1): . \left(kx^2 + \frac{6}{x^2}\right)^4

    Expression (2): . \left(kx^3 + \frac{p}{x^3}\right) ^6

    If we expand those binomials, we find a constant term in the middle (a term with no x's.)


    In expression (1), this term is: . {4\choose2}\left(kx^2\right)^2\left(\frac{6}{x^2}\  right)^2 \:=\:6\cdot k^2\cdot x^4\cdot\frac{36}{x^4} \;=\;216k^2

    In expression (2), this term is: . {6\choose3}\left(kx^3\right)^3\left(\frac{p}{x^3}\  right)^3 \;=\;20\cdot k^3\cdot x^9\cdot\frac{p^3}{x^9} \;=\;20k^3p^3


    Since these terms are equal: . 20k^3p^3 \:=\:216k^2 \quad\Rightarrow\quad \frac{k^3}{k^2} \:=\:\frac{216}{20p^3}


    Therefore: . k \;=\;\frac{54}{5p^3}

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  3. #3
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    Jan 2009
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    Amazing! It actually makes sense now! I didn't realise that you just made them equal to each other and then simplified....

    Thanks
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