15. For the set $A=\{1,2,3, \cdots, n\}$ and each of its non-empty subsets, define a unique “alternating sum” as follows: rearrange the subset in decreasing order, then starting from the largest number, alternately subtract or add the subsequent numbers to get the result. For example, the “alternating sum” of the set $\{1,2,4,7,10\}$ is $10-7+4-2+1=6$, the “alternating sum” of the set $\{7,10\}$ is $10-7=3$, and the “alternating sum” of the set $\{5\}$ is 5, and so on. Find the total of the “alternating sums” of all subsets of $A$.
15. The number of non-empty subsets of set $A$ is $2^{n}-1$. Excluding the set $\{n\}$, there are $2^{n}-2$ non-empty subsets. These can be divided into two categories: the first category consists of subsets containing $n$, and the second category consists of subsets not containing $n$. The number of subsets in these two categories is the same. If $A_{i}$ is in the second category, then $A_{i} \cup\{n\}$ is in the first category, and the sum of their "alternating sums" is $n$. Conversely, if $A_{k}$ is in the first category, then the set formed by removing $n$ from $A_{k}$ is in the second category. Thus, subsets in the first and second categories appear in pairs, each having $\frac{1}{2}\left(2^{n}-2\right)$ subsets. Additionally, the "alternating sum" of $\{n\}$ is $n$. Therefore, the total "alternating sum" is $\frac{1}{2}\left(2^{n}-2\right) \cdot n + n = n \cdot 2^{n-1}$.