2. There are three numbers arranged in sequence: $3, 9, 8$. For any two adjacent numbers, the difference between the right number and the left number is written between these two numbers, resulting in a new sequence $3, 6, 9, -1, 8$, which is called the first operation; after the second similar operation, a new sequence $3, 3, 6, 3, 9, -10, -1, 9, 8$ is produced; continue this operation. Ask: Starting from the sequence 3, 9, 8, what is the sum of all numbers in the new sequence after the 100th operation?
For convenience, let the sequence of $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. According to the problem, the newly added numbers are $a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{n}-a_{n-1}$. Therefore, the sum of the newly added numbers is
$$
\begin{array}{l}
\left(a_{2}-a_{1}\right) + \left(a_{3}-a_{2}\right) + \cdots + \left(a_{n}-a_{n-1}\right) \\
=a_{n}-a_{1} .
\end{array}
$$

The original sequence is three numbers $3, 9, 8$. After the first operation, the resulting sequence is $3, 6, 9, -1, 8$. By equation (1), the sum of the two newly added numbers is
$$
6 + (-1) = 5 = 8 - 3.
$$

After the second operation, by equation (1), the sum of the four newly added numbers is
$$
3 + 3 + (-10) + 9 = 5 = 8 - 3.
$$

Thus, after the 100th operation, the sum of all numbers in the new sequence is
$$
(3 + 9 + 8) + 100 \times (8 - 3) = 520.
$$