5. Try to simplify $\sum_{k=0}^{n} \frac{(-1)^{k} C_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}$ into the form $\frac{p(n)}{q(n)}$, where $p(n)$ and $q(n)$ are two polynomials with integer coefficients.
5. Consider the following:
$$
\begin{array}{l}
k^{3}+9 k^{2}+26 k+24=(k+2)(k+3)(k+4) . \\
\text { i. } S(n)=\sum_{k=0}^{n} \frac{(-1)^{k} C_{k}^{k}}{k^{3}+9 k^{2}+26 k+24} .
\end{array}
$$

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

Let \( T(n)=(n+1)(n+2)(n+3)(n+4) S(n) \)
$$
=\sum_{k=0}^{n}\left[(-1)^{k} C_{n+1}^{k+1}(k+1)\right] \text {. }
$$

For \( n \geqslant 1 \), we have \( \sum_{i=0}^{n}(-1)^{i} C_{n}^{i}=0 \).
Because \( (1-1)^{n}=C_{n}^{n}-C_{n}^{1}+C_{n}^{2}+\cdots+(-1)^{n} C_{n}^{n}=0 \).
$$
\begin{array}{l}
\text { and } \sum_{i=0}^{n}(-1)^{i} C_{n}^{i} i=\sum_{i=1}^{n}(-1)^{i} \frac{i n!}{i!} \\
+(-1) \frac{0 n!}{0!} \\
=\sum_{i=1}^{n}(-1)^{i} \frac{n!}{(i-1)!(n-i)!} \\
=\sum_{i=1}^{n}(-1)^{i} n C_{n-1}^{i-1} \\
=n \sum_{i=1}^{n}(-1)^{c} C_{m i} \\
=-n \sum_{i=1}^{n}(-1)^{i-1} C_{n-1}^{i-1} . \\
\end{array}
$$

Substituting \( i-1 \) with \( j \) in the above expression, and using (*), we get
$$
\sum_{i=0}^{n}(-1) C_{n} i=(-n) \sum_{j=0}^{n-1}(-1)^{j} C_{n-1}^{j}=0 .(* *)
$$

Therefore, \( T(n)=\sum_{k=0}^{n}\left[(-1)^{k} C_{n+4}^{*+4}(k+1)\right] \)
$$
\begin{array}{l}
=\sum_{k=0}^{n}(-1)^{k+4} C^{*+1}(k+1) \\
=\sum_{k=1}^{n}(-1)^{k+4} C_{n+4}^{k+4}(k+1) \\
=\left[-3+2(n+4) \cdots C_{n+4}^{n}\right] .
\end{array}
$$

Substituting \( j=k+4 \) into the above expression,
$$
\begin{aligned}
T(n)= & \sum_{j=-0}^{n+4}(-1)^{j} C_{a+4}^{j}(j-3) \\
& {\left[2 n+8-3-\frac{(n+-1)(n+3)}{2}\right] } \\
= & \sum_{j=0}^{L}(-1)^{j} C_{n+4}^{j} j-3 \sum_{j=0}^{n+4}(-1)^{j} C_{n+4}^{j} \\
& -\frac{1}{2}\left(4 n+10-n^{i}-7 n-12\right) .
\end{aligned}
$$
\(\frac{Z}{3}\)
$$
T(n)=\frac{n^{2}+3 n+2}{2} \text {. }
$$

Thus, \( S(n)=\frac{T(n)}{(n+1)(n+2)(n+3)(n+4)} \)
$$
\begin{array}{l}
=\frac{n+3 n+2}{2(n+1)(n+2)(n+3)(n+1)} \\
=\frac{(n+1)(n+2)}{2(n+1)(n+2)(n+3)(n+4)} \\
=\frac{1}{2(n+3)(n+4)},
\end{array}
$$

i.e., \( \sum_{k=1}^{n} \frac{(-1)^{*} C^{*}}{k^{3}+9 k^{2}+26 k+24}=\frac{1}{2(n-3)(n+4)} \).
(Contributed by Liu Jiangfeng)