We are tasked with finding the number of subsets of $[n] = \{1, 2, \dots, n\}$ whose size is a multiple of 3, and this can be generalized to any natural number $n$. The hint suggests using the third root of unity and the binomial formula.

Step 1: Understanding the Roots of Unity

Let $z$ be a cube root of unity, i.e., a complex number such that $z^3 = 1$. The roots of the equation $z^3 - 1 = 0$ are: $z = 1, \quad \omega = e^{2\pi i / 3}, \quad \omega^2 = e^{4\pi i / 3}.$ These roots satisfy the property: $1 + \omega + \omega^2 = 0.$ We will use these roots to separate subsets whose sizes are multiples of 3.

Step 2: Total Number of Subsets of $[n]$

The total number of subsets of $[n]$ is given by $2^n$, since each element of $[n]$ can either be included or not in a subset.

Step 3: Generating Function for Subsets of Different Sizes

The number of subsets of size $k$ is given by the binomial coefficient $\binom{n}{k}$. To count only those subsets whose size is a multiple of 3, we can use generating functions and roots of unity.

The generating function for the number of subsets of size $k$ is: $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k.$ To isolate the subsets whose sizes are multiples of 3, we will use the cube roots of unity. Specifically, we will compute: $f(1) + f(\omega) + f(\omega^2),$ where $f(x) = (1 + x)^n$. This sum will help us extract only the terms where $k$ is a multiple of 3.

Step 4: Applying Roots of Unity

Now, consider the sum: $S = f(1) + f(\omega) + f(\omega^2) = (1 + 1)^n + (1 + \omega)^n + (1 + \omega^2)^n = 2^n + (1 + \omega)^n + (1 + \omega^2)^n.$ Using the binomial expansion, we can express: $(1 + \omega)^n = \sum_{k=0}^{n} \binom{n}{k} \omega^k, \quad (1 + \omega^2)^n = \sum_{k=0}^{n} \binom{n}{k} \omega^{2k}.$ The key observation is that the powers of $\omega$ cycle every 3 terms, i.e., $\omega^3 = 1$, $\omega^4 = \omega$, and so on. Therefore, adding these up will isolate the terms where $k$ is a multiple of 3.

Step 5: Simplifying the Sum

Because $1 + \omega + \omega^2 = 0$, the terms where $k \equiv 1 \pmod{3}$ and $k \equiv 2 \pmod{3}$ will cancel out when we add $f(1)$, $f(\omega)$, and $f(\omega^2)$. Thus, we are left with only the terms where $k$ is a multiple of 3.

Thus, the number of subsets of $[n]$ whose size is a multiple of 3 is: $\frac{f(1) + f(\omega) + f(\omega^2)}{3} = \frac{2^n + (1 + \omega)^n + (1 + \omega^2)^n}{3}.$

Step 6: Special Case for $n$ Divisible by 6

If $n$ is divisible by 6, the expression simplifies further due to symmetries in powers of $\omega$. However, the general formula works for any $n$.

Final Answer:

The number of subsets of $[n]$ whose size is a multiple of 3 is: $\frac{2^n + (1 + \omega)^n + (1 + \omega^2)^n}{3}.$

Would you like to go deeper into the roots of unity or explore specific cases? Here are a few follow-up questions for further insight:

1. What is the value of $\omega^k$ for various $k$ values?
2. Can this method be generalized to find subsets whose size is a multiple of any number $m$?
3. How does the formula change when $n$ is not divisible by 6?
4. How does the inclusion-exclusion principle relate to this counting method?
5. What is the computational complexity of evaluating this sum for large $n$?

Tip: Roots of unity are useful not only in combinatorics but also in solving polynomial equations efficiently.