$n$ 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, $b(n)-a(n)=\binom{n}{3}$.
