$3 \cdot 33$ Find the sum $\sum_{n=1}^{1994}(-1)^{n} \frac{n^{2}+n+1}{n!}$.
［Solution］Since
$$
\frac{n^{2}+n+1}{n!}=\frac{n}{(n-1)!}+\frac{n+1}{n!}
$$

where $n \in N, 0!=1$. Therefore,
$$
\begin{aligned}
& \sum_{n=1}^{1994}(-1)^{n} \cdot \frac{n^{2}+n+1}{n!} \\
= & \sum_{n=1}^{1994}(-1)^{n}\left(\frac{n}{(n-1)!}+\frac{n+1}{n!}\right) \\
= & -\frac{1}{(1-1)!}+\frac{1994+1}{1994!} \\
= & -1+\frac{1995}{1994!} .
\end{aligned}
$$