Combinatorial sum!

$\displaystyle \sum_{k=0}^{18}\frac{(-1)^k{18\choose k}}{k^3+9k^2+26k+24} = \sum_{k=0}^{18}\frac{(-1)^k{18\choose k}}{(k+2)(k+3)(k+4)}$

To evaluate this sum, we will use Integration.
We know,
$\displaystyle (1+x)^n = \sum_{k=0}^n{n\choose k} x^k$

Multiplying both sides by $x$ , we get,
$\displaystyle x(1+x)^n = \sum_{k=0}^n{n\choose k} x^{k+1}$

Integrating both sides, with limits as $0$ to $x$ , we have,
$\displaystyle\int_0^x x(1+x)^n dx = \sum_{k=0}^n\int_0^x{n\choose k} x^{k+1} dx$

Using IBP, and putting the limits, we get,
$\displaystyle\frac{x(1+x)^{n+1}}{n+1}-\frac{(1+x)^{n+2}}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \sum_{k=0}^n{n\choose k} \frac{x^{k+2}}{k+2}$

Integrating 2 more times with the same limits, we get,
$\displaystyle\frac{2(n+4)(2x+x(1+x)^{n+3})+3-3(1+x)^{n+4}+x^2(n+3)(n+4)}{2(n+1)(n+2)(n+3)(n+4)} = \sum_{k=0}^n{n\choose k} \frac{x^{k+4}}{(k+2)(k+3)(k+4)}$

Substituting $x=-1$ and $n=18$ ,
$\displaystyle\sum_{k=0}^n{n\choose k} \frac{(-1)^k}{(k+2)(k+3)(k+4)} = \frac{380}{2\times19\times20\times21\times22} =\frac{1}{924} = \frac{1}{n}$

Therefore,
$\boxed{n=924}$

nice question