Number of pieces $\frac{n^3+5n+6}{6}$

per Peter C. Heinig

Define a number of planes in space to be in general arrangement when

1) no two planes are parallel,

2) there are no two parallel intersection lines,

3) there is no point common to four or more planes.

Suppose there are already n-1 planes in general arrangement, thus defining the maximal number of regions in space obtainable by n-1 planes and now one more plane is added in general arrangement. Then it will cut each of the n-1 planes and acquire intersection lines which are in general arrangement.

These lines on the new plane define the maximal number of regions in 2-space definable by n-1 straight lines, hence this is (Lazy Caterer's sequence)(n-1).

Each of this regions acts as a dividing wall, thereby creating as many new regions in addition to the a(n-1) regions already there, hence a(n)=a(n-1)+(Lazy Caterer's sequence)(n-1).

So,
f(n) = ${n \choose 3} + {n \choose 2} + {n \choose 1} + {n \choose 0}$ = $\frac{1}{6}(n^3+5n+6)$