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

1. ## 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.

2. 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.

3. Your answer is the simplest form I guess. Is this the same?:
$\frac{(n - 14)2^{n + 4} + 2n^3 + 25n^2 + 117n + 214}{(n + 1)(n + 2)(n + 3)(n + 4)}$