Problem 1534

1. Define the problem and notation:
We need to determine the sum
$$\sum_{\substack{S \subset \{1,2,\ldots,3n\} \\ 3 \mid f(S)}} f(S)$$
where $f(S)$ is the sum of the elements of $S$ and $S$ runs through all subsets of $\{1, 2, \ldots, 3n\}$ such that $f(S)$ is a multiple of 3.

2. Sum of all elements:
The sum of all elements in $\{1, 2, \ldots, 3n\}$ is:
$$\sum_{k=1}^{3n} k = \frac{3n(3n+1)}{2}$$
This sum is divisible by 3 because $3n(3n+1)/2$ is an integer and $3n$ is a multiple of 3.

3. Complementary subsets:
For any subset $S \subset \{1, 2, \ldots, 3n\}$, let $S^c$ be its complement. Then:
$$f(S) + f(S^c) = \frac{3n(3n+1)}{2}$$
Since $\frac{3n(3n+1)}{2}$ is divisible by 3, $f(S)$ is a multiple of 3 if and only if $f(S^c)$ is a multiple of 3.

4. **Sum of $f(S)$ and $f(S^c)$:**
$$U = \sum_{\substack{S \subset \{1,2,\ldots,3n\} \\ 3 \mid f(S)}} f(S)$$
By symmetry, we have:
$$U = \sum_{\substack{S \subset \{1,2,\ldots,3n\} \\ 3 \mid f(S)}} f(S^c)$$
Therefore:
$$2U = \sum_{\substack{S \subset \{1,2,\ldots,3n\} \\ 3 \mid f(S)}} (f(S) + f(S^c)) = \frac{3n(3n+1)}{2} \sum_{\substack{S \subset \{1,2,\ldots,3n\} \\ 3 \mid f(S)}} 1$$

5. **Number of subsets $S$ such that $3 \mid f(S)$:**
Let $a_n$ denote the number of subsets $S \subset \{1, 2, \ldots, 3n\}$ such that $3 \mid f(S)$. Then:
$$2U = \frac{3n(3n+1)}{2} a_n$$
$$U = \frac{3n(3n+1)}{4} a_n$$

6. **Recursive relation for $a_n$:**
Consider any subset $S' \subset \{1, 2, \ldots, 3n-3\}$. We can adjoin one set from one of the following pairs to it:
- $\{3n\}$ or $\{3n-1, 3n-2\}$ if $3 \mid f(S')$
- $\{3n-1\}$ or $\{3n, 3n-1\}$ if $3 \mid f(S') + 1$
- $\{3n-2\}$ or $\{3n, 3n-2\}$ if $3 \mid f(S') + 2$

This gives us $2 \times 2^{3n-3} = 2^{3n-2}$ sets $S \subset \{1, 2, \ldots, 3n\}$ with $3 \mid f(S)$.

7. Counting subsets:
Any set not generated by this method must contain all of $\{3n, 3n-1, 3n-2\}$ or none of it, and in either case, it is counted in $a_{n-1}$. Hence:
$$a_n = 2a_{n-1} + 2^{3n-2}$$

8. Solving the recurrence:
Substitute $b_n = \frac{a_n}{2^n}$:
$$b_n = b_{n-1} + 4^{n-1}$$
With $a_1 = 4$, we have $b_1 = 2$. Solving the recurrence:
$$b_n = 2 + 4 + 4^2 + \cdots + 4^{n-1} = \frac{4^n - 1}{3} + 2 = \frac{4^n + 2}{3}$$
Therefore:
$$a_n = 2^n b_n = 2^n \left( \frac{4^n + 2}{3} \right) = \frac{2^{n+1}(2^{2n-1} + 1)}{3}$$

9. **Final expression for $U$:**
$$U = \frac{3n(3n+1)}{4} a_n = \frac{3n(3n+1)}{4} \cdot \frac{2^{n+1}(2^{2n-1} + 1)}{3} = 2^{n-1}(2^{2n-1} + 1)n(3n+1)$$