Dividing the plane

Prove:

Variant
Prove that \(n\) lines can divide the plane into at most \(1+\frac{1}{2}(n^2+n)\) regions.
Solution
Let \(P(n)=1+\frac{1}{2}(n^2+n)\)

For \(n=1\): a single line divides the plane into two regions, and \(P(1)=1+\frac{1}{2}(1^2+1)=2\).

Induction step \(n-1 \to n\):
A newly added line intersects \(n-1\) existing lines and is divided by them into \(n\) parts (two rays and \(n-2\) line segments). Each of these parts divides some part of the plane into two new ones, so

\(P(n)=P(n-1)+n=1+\frac{1}{2}\left((n-1)^2+(n-1)\right)=\\ =1+\frac{1}{2}\left(n^2-2n+1+n-1+2n\right)= 1+\frac{1}{2}(n^2+n)\)
Variant
Try to derive a similar upper bound for dividing three-dimensional space by a number of planes.
Solution
Let \(S(n)\) be the number of regions we are seeking.

We can see that \(S(1)=2, S(2)=4, S(3)=8, S(4)=15\), etc.

A newly added plane will intersect \(n-1\) existing planes and will be divided by them into at most \(P(n-1)=1+\frac{1}{2}(n^2-n)\) planar regions. Each such region divides some region of space into two new regions, so

\( S(n)=S(n-1)+1+\frac{1}{2}(n^2-n)= 1+\sum\limits_{i=1}^n\left(1+\frac{1}{2}(n^2-n)\right)= n+1+\frac{1}{2}\sum\limits_{i=1}^n i^2 -\frac{1}{2}\sum\limits_{i=1}^n i\),

which we can further simplify using equations for the sum of series.
Answer
Using \(n\) planes we can divide three-dimensional space into at most \(\displaystyle \frac16 n^3 + \frac56 n + 1 \) regions.