$3 \cdot 28$ Try to express $\sum_{k=0}^{n} \frac{(-1)^{k} C_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}$ in the form $\frac{p(n)}{q(n)}$, where $p(n)$ and $q(n)$ are two polynomials with integer coefficients.
[Solution] Factorization yields
$$k^{3}+9 k^{2}+26 k+24=(k+2)(k+3)(k+4)$$

Let
$$S(n)=\sum_{k=0}^{n} \frac{(-1)^{k} \cdot C_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}$$

Then $\square$

Let
$$\begin{aligned}
S(n)= & \sum_{k=0}^{n} \frac{(-1)^{k} \cdot n!}{k!(n-k)!(k+2)(k+3)(k+4)} \\
= & \sum_{k=0}^{n}\left[\frac{(-1)^{k}(n+4)!}{(k+4)!(n-k)!}\right] \\
& {\left[\frac{k+1}{(n+1)(n+2)(n+3)(n+4)}\right] }
\end{aligned}$$

Then $\square$
$$\begin{array}{l}
T(n)=(n+1)(n+2)(n+3)(n+4) \cdot S(n) \\
T(n)=\sum_{k=0}^{n}\left[(-1)^{k} C_{n+4}^{k+4} \cdot(k+1)\right]
\end{array}$$

When $n \geqslant 1$, since $(1-1)^{n}=0$, we have
$$\sum_{i=0}^{n}(-1)^{i} C_{n}^{i}=0$$

And
$$\begin{aligned}
\sum_{i=0}^{n}(-1)^{i} C_{n}^{i} \cdot i & =\sum_{i=1}^{n}(-1)^{i} \cdot \frac{n!\cdot i}{i!(n-i)!} \\
& =\sum_{i=1}^{n}(-1)^{i} \frac{n!}{(i-1)!(n-i)!} \\
& =n \sum_{i=1}^{n}(-1)^{i} \frac{(n-1)!}{(i-1)!(n-i)!} \\
& =n \sum_{i=1}^{n}(-1)^{i} C_{n-1}^{i-1} \\
& =-n \sum_{i=1}^{n}(-1)^{i-1} \cdot C_{n-1}^{i-1}
\end{aligned}$$

Let $j=i-1$, then from the above and (1), when $n \geqslant 2$,
$$\sum_{i=0}^{n}(-1)^{i} C_{n}^{i} \cdot i=(-n) \cdot \sum_{j=0}^{n-1}(-1)^{j} C_{n-1}^{j}=0$$

Therefore
$$\begin{aligned}
T(n) & =\sum_{k=0}^{n}\left[(-1)^{k} C_{n+4}^{k+4}(k+1)\right] \\
& =\sum_{k=0}^{n}\left[(-1)^{k+4} C_{n+4}^{k+4}(k+1)\right] \\
& =\sum_{k=-4}^{n}(-1)^{k+4} C_{n+4}^{k+4}(k+1)-\left[-3+2(n+4)-C_{n+4}^{2}\right]
\end{aligned}$$

Substituting $j=k+4$ into the above,
$$\begin{aligned}
T(n) & =\sum_{j=0}^{n+4}(-1)^{j} C_{n+4}^{j}(j-3)-\left[2 n+5-\frac{(n+4)(n+3)}{2}\right] \\
& =\sum_{j=0}^{n+4}(-1)^{j} C_{n+4}^{j} j-3 \sum_{j=0}^{n+4}(-1)^{j} C_{n+4}^{j}+\frac{1}{2}\left(n^{2}+3 n+2\right)
\end{aligned}$$

From (1) and (2), the first two terms on the right side of the above equation are both 0, so
$$\begin{aligned}
T(n) & =\frac{1}{2}\left(n^{2}+3 n+2\right) \\
& =\frac{1}{2}(n+1)(n+2) \\
S(n) & =\frac{T(n)}{(n+1)(n+2)(n+3)(n+4)} \\
& =\frac{1}{2(n+3)(n+4)}
\end{aligned}$$

That is, $\sum_{k=0}^{n} \frac{(-1)^{k} C_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}=\frac{1}{2(n+3)(n+4)}$.