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Math Help - Write the sum in the form p(n)/q(n)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Write the sum in the form p(n)/q(n)?

    Write the sum \sum_{k = 0}^{n} \left\{\frac{(-1)^k\cdot {n\choose k}}{k^3 + 9k^2 + 26k + 24}\right\}, in the form \left\{\frac{p(n)}{q(n)}\right\}, where p and q are polynomials with integer coefficients.
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Write the sum \sum_{k = 0}^{n} \left\{\frac{(-1)^k\cdot {n\choose k}}{k^3 + 9k^2 + 26k + 24}\right\}, in the form \left\{\frac{p(n)}{q(n)}\right\}, where p and q are polynomials with integer coefficients.
    A bit of experimentation with n=1,2,3 suggests that the answer should be \frac1{2(n+3)(n+4)}. Also note that k^3 + 9k^2 + 26k + 24 = (k+2)(k+3)(k+4). So \frac{{n\choose k}}{k^3 + 9k^2 + 26k + 24} = \frac{k+1}{(n+1)(n+2)(n+3)(n+4)}{n+4\choose k+4}.

    With that motivation, start from the identity (1+x)^{n+4} - 1 - (n+4)x - \tfrac12(n+4)(n+3)x^2 - \tfrac16(n+4)(n+3)(n+2)x^3 = \sum_{k=0}^n{n+4\choose k+4}x^{k+4}, divide both sides by x^3, differentiate, and put x=-1. The result simplifies to \sum_{k = 0}^{n} \frac{(-1)^k\cdot {n\choose k}}{k^3 + 9k^2 + 26k + 24} = \frac1{2(n+3)(n+4)}, as expected.
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  3. #3
    Super Member fardeen_gen's Avatar
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    Your answer is the simplest form I guess. Is this the same?:
    \frac{(n - 14)2^{n + 4} + 2n^3 + 25n^2 + 117n + 214}{(n + 1)(n + 2)(n + 3)(n + 4)}
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