1.48 In the expression $x_{1}: x_{2}: \cdots: x_{n}$, use parentheses to indicate the order of operations, and the result can be written in fractional form:
$$\frac{x_{i_{1}} x_{i_{2}} \cdots x_{i_{k}}}{x_{j_{j_{1}}}^{x_{j_{2}}} \cdots x_{j_{n-k}}}$$
(At the same time, each letter in $x_{1}, x_{2}, \cdots, x_{n}$ may appear in the numerator or in the denominator.) How many different fractions can be obtained by adding parentheses in all possible ways?
[Solution] Clearly, $x_{1}$ will be in the numerator of the resulting fraction.
When parentheses are added arbitrarily, the division sign before $x_{2}$ or the expression involving $x_{2}$ will be in the denominator, so $x_{2}$ will always be in the denominator.

We will use induction to prove that each of the remaining $n-2$ letters $x_{3}, x_{4}, \cdots, x_{n}$ can independently be in the numerator or the denominator, thus yielding $2^{n-2}$ fractions.

In fact, when $n=3$, we can get 2 fractions:
$$\left(x_{1}: x_{2}\right): x_{3}=\frac{x_{1}}{x_{2} x_{3}} \text { and } x_{1}:\left(x_{2}: x_{3}\right)=\frac{x_{1} x_{3}}{x_{2}} \text {. }$$

So the conclusion holds for $n=3$.
Assume the conclusion holds for $n=k$, we will prove it for $n=k+1$.
Suppose that after adding parentheses in some way, the expression $x_{1}: x_{2}: \cdots: x_{k}$ can be written in the form of some fraction $A$. If in this expression, $x_{k}: x_{k+1}$ replaces $x_{k}$, then $x_{k}$ will still be in its original position in $A$, and $x_{k+1}$ will not be in the original position of $x_{k}$, that is, if $x_{k}$ was originally in the denominator, then $x_{k+1}$ will be in the numerator, and vice versa.

On the other hand, after adding parentheses in the fraction $A$, there will certainly be an expression of the form $\left(p: x_{k}\right)$, where $P$ is either the letter $x_{k-1}$ or some parentheses. Replacing $\left(p: x_{k}\right)$ with $\left.\left[P: x_{k}\right): x_{k+1}\right]=P$ $:\left(x_{k} \cdot x_{k+1}\right)$, it is clear that in the original fraction $A$, $x_{k} \cdot x_{k+1}$ replaces $x_{k}$. That is, $x_{k+1}$ can be in the same position as $x_{k}$, either both in the numerator or both in the denominator.

By the principle of mathematical induction, each of the remaining $n-2$ letters can independently be in the numerator or the denominator.

Therefore, by adding parentheses in all possible ways, we can obtain $2^{n-2}$ different fractions.