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

**fardeen_gen** · April 29th 2009, 05:52 AM

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.

**Opalg** · April 30th 2009, 11:37 AM

Originally Posted by fardeen_gen

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.

**fardeen_gen** · April 30th 2009, 07:48 PM

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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