Let $S$ be the set $\{1, 2, ..., 10\}$. Let $A$ be a subset of $S$.
We arrange the elements of $A$ in increasing order, that is, $A = \{a_1, a_2, ...., a_k\}$ with $a_1 < a_2 < ... < a_k$. 
Define [i]WSUM [/i] for this subset as $3(a_1 + a_3 +..) + 2(a_2 + a_4 +...)$ where the first term contains the odd numbered terms and the second the even numbered terms. 
(For example, if $A = \{2, 5, 7, 8\}$, [i]WSUM [/i] is $3(2 + 7) + 2(5 + 8)$.) 
Find the sum of [i]WSUMs[/i] over all the subsets of S. 
(Assume that WSUM for the null set is $0$.)
1. **Define the problem in general terms:**
   Let \( S = \{1, 2, \ldots, n\} \). For a subset \( A \) of \( S \), arranged in increasing order \( A = \{a_1, a_2, \ldots, a_k\} \) with \( a_1 < a_2 < \ldots < a_k \), define the *WSUM* of \( A \) as:
   \[
   \text{WSUM}(A) = 3(a_1 + a_3 + \ldots) + 2(a_2 + a_4 + \ldots)
   \]
   We need to find the sum of *WSUM*s over all subsets of \( S \).

2. **Generalize the problem:**
   Define the *WUMBO* of a subset \( A \) as:
   \[
   \text{WUMBO}(A) = f(a_1) + g(a_2) + f(a_3) + g(a_4) + \cdots
   \]
   For *WSUM*, set \( f(m) = 3m \) and \( g(m) = 2m \).

3. **Calculate the total sum \( T \) of all *WUMBO*s over all subsets of \( S \):**
   Consider an element \( m \) in \( S \). The number of ways for \( m \) to appear as the \( i \)-th term in a subset \( A \) depends on whether \( i \) is odd or even.

4. **Count the occurrences of \( f(m) \) and \( g(m) \):**
   - For \( m > 1 \):
     - The number of ways for \( m \) to appear as an odd-indexed term is the number of subsets of \( \{1, 2, \ldots, m-1\} \) with an even number of elements, times the number of subsets of \( \{m+1, m+2, \ldots, n\} \). This is:
       \[
       2^{m-2} \cdot 2^{n-m} = 2^{n-2}
       \]
     - Similarly, the number of ways for \( m \) to appear as an even-indexed term is:
       \[
       2^{m-2} \cdot 2^{n-m} = 2^{n-2}
       \]
   - For \( m = 1 \):
     - \( 1 \) always appears as the first term, so \( f(1) \) appears \( 2^{n-1} \) times, and \( g(1) \) never appears.

5. **Sum the contributions:**
   - The total contribution of \( f(m) \) and \( g(m) \) for \( m > 1 \) is:
     \[
     2^{n-2} \cdot (f(m) + g(m))
     \]
   - For \( m = 1 \):
     \[
     2^{n-1} \cdot f(1)
     \]

6. **Combine the results:**
   \[
   T = 2^{n-1} f(1) + \sum_{m=2}^n 2^{n-2} (f(m) + g(m))
   \]

7. **Substitute \( f(m) = 3m \) and \( g(m) = 2m \):**
   \[
   T = 2^{n-1} \cdot 3 \cdot 1 + \sum_{m=2}^n 2^{n-2} (3m + 2m)
   \]
   \[
   T = 2^{n-1} \cdot 3 + 2^{n-2} \sum_{m=2}^n 5m
   \]

8. **Simplify the sum:**
   The sum of the first \( n \) natural numbers is:
   \[
   \sum_{m=1}^n m = \frac{n(n+1)}{2}
   \]
   Therefore:
   \[
   \sum_{m=2}^n m = \frac{n(n+1)}{2} - 1
   \]
   \[
   T = 2^{n-1} \cdot 3 + 2^{n-2} \cdot 5 \left( \frac{n(n+1)}{2} - 1 \right)
   \]
   \[
   T = 2^{n-1} \cdot 3 + 2^{n-2} \cdot 5 \left( \frac{n(n+1) - 2}{2} \right)
   \]
   \[
   T = 2^{n-1} \cdot 3 + 2^{n-3} \cdot 5 (n(n+1) - 2)
   \]
   \[
   T = 2^{n-1} \cdot 3 + 2^{n-3} \cdot 5 (n^2 + n - 2)
   \]
   \[
   T = 2^{n-1} \cdot 3 + 2^{n-3} \cdot 5n^2 + 2^{n-3} \cdot 5n - 2^{n-3} \cdot 10
   \]
   \[
   T = 2^{n-1} \cdot 3 + 2^{n-3} \cdot 5n^2 + 2^{n-3} \cdot 5n - 2^{n-2} \cdot 5
   \]
   \[
   T = 2^{n-3} (5n^2 + 5n + 2)
   \]

The final answer is \( \boxed{2^{n-3} (5n^2 + 5n + 2)} \).