Difference between lines dividing planes and planes dividing space Let a(n) represent the number of regions that the plane R2 is broken into by n lines (no 2 of which are parallel, and no 3 of which intersect in a single point). 
Let b(n) represent the number of regions that the 3-dimensional space R3 is broken into by n planes (no 2 of which are parallel, and no 3 of which intersect in a shared line).  
Find either a recursive formula or a closed-form solution for a(n) and b(n). 
So I know that a(n)=C(n+1,2)+1. Why should b(n) be any different? I simply cannot visualize it. 
 A: $n$ points divides $\mathbb{R}^1$ into at most $\binom{n}{0}+\binom{n}{1}$ pieces.

When dividing $\mathbb{R}^2$, line $n$ meets the $n-1$ previous lines at $n-1$ points, dividing line $n$ into $\binom{n-1}{0}+\binom{n-1}{1}$ pieces. Each of those pieces of line $n$ divides a piece of $\mathbb{R}^2$ in two, adding $\binom{n-1}{0}+\binom{n-1}{1}$ pieces of $\mathbb{R}^2$. We start with $1=\binom{n}{0}$ piece of $\mathbb{R}^2$, so after adding $n$ lines, we get
$$
\begin{align}
a(n)
&=\binom{n}{0}+\sum_{k=1}^n\left[\binom{k-1}{0}+\binom{k-1}{1}\right]\\
&=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}
\end{align}
$$
pieces of $\mathbb{R}^2$. We used the Hockey Stick Identity to evaluate the summations above.

When dividing $\mathbb{R}^3$, plane $n$ meets the $n-1$ previous planes at $n-1$ lines, dividing plane $n$ into $\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}$ pieces. Each of those pieces of plane $n$ divides a piece of $\mathbb{R}^3$ in two, adding $\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}$ pieces of $\mathbb{R}^3$. We start with $1=\binom{n}{0}$ piece of $\mathbb{R}^3$, so after adding $n$ planes, we get
$$
\begin{align}
b(n)
&=\binom{n}{0}+\sum_{k=1}^n\left[\binom{k-1}{0}+\binom{k-1}{1}+\binom{k-1}{2}\right]\\
&=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}
\end{align}
$$
pieces of $\mathbb{R}^3$.

Thus, $b(n)-a(n)=\binom{n}{3}$.
A: $a(n)$ is known as the lazy caterer's sequence, while $b(n)$ is known as the cake number. They are different starting from $n=3$, notice placing the three "canonical planes" split space into $8$ quadrants, while $a(n)=7$
