Three, (20 points) When the faces of an $n(n \geqslant 3)$-sided pyramid are extended into planes, into how many parts can space be divided at most?
Three, let the $n$ side planes divide the space into $a_{n}$ parts. Since the $n$ planes intersect at one point, the $n$-th plane intersects with the previous $n-1$ planes, resulting in $n-1$ intersection lines. These $n-1$ intersection lines, intersecting at one point, divide the $n$-th plane into $2(n-1)$ parts, and each part divides the space into two. Therefore,
$$
\begin{array}{l}
a_{n}=a_{n-1}+2(n-1), a_{1}=2 . \\
\text { Hence } a_{n}=\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right)+a_{1} \\
=2(n-1)+2(n-2)+\cdots+2 \times 1+2 \\
=n(n-1)+2 .
\end{array}
$$

The base plane intersects with the $n$ side planes, resulting in $n$ intersection lines. Since the base plane is a convex $n$-gon, no three of these $n$ intersection lines intersect at the same point.

To maximize the number of parts the plane divides the space into, no two of these $n$ intersection lines should be parallel. Let these $n$ intersection lines divide the base plane into the maximum number of parts, denoted as $b_{n}$. Since the $n$-th intersection line intersects with the other $n-1$ lines, resulting in $n-1$ intersection points, these $n-1$ points divide the $n$-th line into $n$ segments, and each segment divides the region it is in into two, so, $b_{n}=b_{n-1}+n, b_{1}=2$. Hence
$$
\begin{array}{l}
b_{n}=\left(b_{n}-b_{n-1}\right)+\left(b_{n-1}-b_{n-2}\right)+\cdots+\left(b_{2}-b_{1}\right)+b_{1} \\
=n+(n-1)+\cdots+2+2 \\
=\frac{n(n+1)}{2}+1 .
\end{array}
$$

Each part of the base plane divides the space it is in into two, so the maximum number of parts the space can be divided into is
$$
\begin{array}{l}
a_{n}+b_{n}=n(n-1)+2+1+\frac{n(n+1)}{2} \\
=\frac{3}{2} n^{2}-\frac{1}{2} n+3(n \geqslant 3)
\end{array}
$$

parts.