Yeah, a Telescoping Sum..

We will prove a more general case.

We have $\displaystyle \sum_{k=0}^{n} \dfrac{(-1)^{k}}{(k+2)(k+3)(k+4)}\binom{n}{k}$ $=\sum_{k=0}^{n} (-1)^{k}\dfrac{k+1}{(n+1)(n+2)(n+3)(n+4)}\binom{n+4}{k+4}$ $=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\sum_{k=4}^{n+4} (-1)^{k}(k-3)\binom{n+4}{k}$

So now we only have to calculate $\sum_{k=4}^{n+4} (-1)^{k}(k-3)\binom{n+4}{k}$

Note that $\sum_{k=4}^{n+4}(-1)^{k}(k-3)\binom{n+4}{k}=\sum_{k=4}^{n+4}(-1)^{k} k\binom{n+4}{k}-3\sum_{k=4}^{n+4}(-1)^{k}\binom{n+4}{k}$ $=\dfrac{1}{n+4}\sum_{k=1}^{n+4}(-1)^{k}\binom{n+3}{k-1}=0$

Thus,

$\sum_{k=4}^{n+4}(-1)^{k}(k-3)\binom{n+4}{k}=-\sum_{k=0}^{3}(-1)^{k}(k-3)\binom{n+4}{k}$ $=\dfrac{(n+1)(n+2)}{2}$

And the sum we seek is then $\dfrac{1}{(n+1)(n+2)(n+3)(n+4)} \times \dfrac{(n+1)(n+2)}{2}=\dfrac{1}{2(n+3)(n+4)}$

Substituting $n=997$ , we will arrive with the answer $\dfrac{1}{2002000}$ , and $m+n=\boxed{2002001}$