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Math Help - binomial thm

  1. #1
    Member maybeline9216's Avatar
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    binomial thm

    Write down the first three terms in the expansion, in descending powers of x, of [ax-(b/x^3)]^8. Hence evaluate the term independent of x of [5x-(1/x^3)]^8.

    I dunno how to do the Hence part, making use of the terms in the expansion.
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  2. #2
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    Quote Originally Posted by maybeline9216 View Post
    Hence evaluate the term independent of x [/B]of [5x-(1/x^3)]^8.
    In this expansion we have terms of form: {8\choose k} (-1)^k(5x)^k \left( \frac{1}{x^3} \right)^{8-k}

    To have a term independent of x we require that the x in numerator cancels out the x in denominator. The exponent of x in numberator is k and the exponent of x in denominator is 3(8-k). We want that k = 3(8-k) so that they cancel out. Therefore, k=6 if we solve this equation.

    Therefore the indepentent term is: {8\choose 6} (-1)^k 5^6
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  3. #3
    Member maybeline9216's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    In this expansion we have terms of form: {8\choose k} (-1)^k(5x)^k \left( \frac{1}{x^3} \right)^{8-k}

    To have a term independent of x we require that the x in numerator cancels out the x in denominator. The exponent of x in numberator is k and the exponent of x in denominator is 3(8-k). We want that k = 3(8-k) so that they cancel out. Therefore, k=6 if we solve this equation.

    Therefore the indepentent term is: {8\choose 6} (-1)^k 5^6
    Umm....did u make use of the first part of the qn?? Cause they specify "Hence" but anyway thks
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