Let $R_{1}, R_{2}, \cdots$, be a family of finite sequences of positive integers defined by the following rules:
$$
R_{1}=\text { (1); }
$$

If $R_{4} =\left(x_{1}, \cdots, x_{s}\right)$, then
$$
R_{n}=\left(1,2, \cdots, x_{1}, 1,2, \cdots, x_{2}, \cdots, 1,2, \cdots, x_{s}\right. \text {, }
$$
$n)$.
For example: $R_{2}=(1,2), R_{3}=(1,1,2,3), R_{4}=(1,1$, $1,2,1,2,3,4)$,
(a) Find the sum of all numbers in $R_{n}$;
(b) Prove: If $n>1$, then the $k$-th term from the left in $R_{n}$ is equal to 1 if and only if the $k$-th term from the right in $R_{n}$ is not equal to 1.
Solution: (a) The easiest way to see that the sum in (a) is $2^{n}-1$ using the knowledge of the pseudo-Pascal triangle is as follows: write out the pseudo-Pascal triangle, and in the sequence, replace the sequence itself with the sum of the numbers, which is represented as follows

By adding a string of 1s (as shown above), it can be extended to the usual Pascal triangle, thus yielding the result.

There are many other ways to prove this. For example, it is easy to use mathematical induction to show and prove that $R_{n}$ contains $2^{n-1}$ ones (exactly half), $2^{n-2}$ twos, and so on, with $R_{n}$ containing exactly one $(n-1)$ and one $n$. It is easy to calculate the sum of all the numbers.