10,11  [ Sums of numerical sequences and difference series ]

How many parts can $n$ planes divide space into?

(Any three planes intersect at one point, no four planes have a common point.)

#
Let's denote this number as $K_{n}$. Let's see how this number changes when the $(n+1)$-th plane is drawn. This plane intersects the other planes along $n$ lines of "general position" and is divided by them into $1 / 2\left(n^{2}+n+2\right)$ pieces (see problem $\underline{60323}$). Each of these pieces divides one of the "old" parts of space into two. Therefore, $K_{n+1}=K_{n}+1 / 2\left(n^{2}+n+2\right)$. It is more convenient to write this formula as $K_{n}=K_{n-1}+1 / 2\left(n^{2}-n+2\right)$.

Since $K_{0}=1$, we have

$2 K_{n}=2+\left(1^{2}-1+2\right)+\left(2^{2}-2+2\right)+\ldots+\left(n^{2}-n+2\right)=\left(1^{2}+2^{2}+\ldots+n^{2}\right)-(1+2+\ldots+n)+2(n+1)=$

$=1 / 6 n(n+1)(2 n+1)-1 / 2 n(n+1)+2(n+1)=1 / 2(n+1)\left(n^{2}-n+6\right)$.

(For the formula for the sum of squares, see problem $\underline{60282}$.)

## Answer

Into $1 / 6\left(n^{3}+5 n+6\right)$ parts.