\section*{Problem 5 - 151245}

It is well known that:

For every natural number \(n\), a plane is divided into exactly \(\frac{1}{2}\left(n^{2}+n+2\right)\) parts by \(n\) lines, no three of which pass through the same point and no two of which are parallel.

Determine for every natural number \(n\) the number of parts into which space is divided by \(n\) planes, no four of which pass through the same point, no three of which have parallel or coincident intersection lines, and no two of which are parallel.
Let \(A_{n} (n=0,1,2, \ldots)\) denote the number of parts into which space is divided by the \(n\) planes (under the given conditions). Clearly, \(A_{0}=1 \quad(*)\).

For \(n>0\), let \(E_{1}, \ldots, E_{n}\) be planes that satisfy the conditions of the problem. Then the plane \(E_{n}\) intersects the planes \(E_{1}, \ldots, E_{n-1}\) in \(n-1\) lines, no three of which pass through the same point and no two of which are parallel.

These lines divide the plane \(E_{n}\) into \(\frac{1}{2}\left(n^{2}-n+2\right)\) parts according to the preliminary remark. Each of these planar parts divides exactly one of the previously existing \(A_{n-1}\) spatial parts into exactly two spatial parts; all other (previously existing spatial parts) remain unchanged. Therefore, we have:

\[
A_{n}=A_{n-1}+\frac{1}{2}\left(n^{2}-n+2\right) \quad(* *)
\]

for \(n=1,2, \ldots\) From \(\left({ }^{*}\right)\) and \(\left({ }^{* *}\right)\), we obtain by induction:

\[
A_{n}=1+\frac{1}{2} \sum_{k=1}^{n}\left(k^{2}-k+2\right)=1+\frac{1}{2} \sum_{k=1}^{n} k^{2}-\frac{1}{2} \sum_{k=1}^{n} k+n
\]

It is well known that:

\[
\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6} \quad \text { and } \quad \sum_{k=1}^{n} k=\frac{n(n+1)}{2}
\]

Thus, we have:

\[
\begin{aligned}
A_{n} & =\frac{12+n(n+1)(2 n+1)-3 n(n+1)+12 n}{12} \\
& =\frac{12(n+1)+(n+1)\left(2 n^{2}+n-3 n\right)}{12}=\frac{1}{6}\left(n^{3}+5 n+6\right)
\end{aligned}
\]

Q.E.D.

\section*{Adapted from \([3]\)}