Variance of the sums of all combinations of a set of numbers Let's assume a finite set of $n$ real numbers: $$\mathbb{V}=\{a,b,c,...,z\}$$ Now we take all the possible combinations of this set, including $0$: $$\mathbb{C}=\{\{0\},\{a\},\{b\},...\{a,b\},\{a,c\},...,\{a,b,c,...,z\}\}$$ After that we sum up all of those combinations: $$\mathbb{S}=\{\{0\},\{a\},\{b\},...\{a+b\},\{a+c\},...,\{a+b+c+...+z\}\}$$ We now have $2^n$ sums in $\mathbb{S}$.
My question
What is the variance of $\mathbb{S}$?
What have I found out and tried so far?
I found out that the mean of $\mathbb{S}$ is the sum of all numbers of $\mathbb{V}$ divided by $2$ and that the variance could be approximated by $n$ times the variance of $1/n$ times the sum of all positive numbers of $\mathbb{V}$ and $1/n$ times the sum of all negative numbers of $\mathbb{V}$. I came up with this approximation because the resulting structure $\mathbb{S}$ has some resemblance to a binomial tree and the variance is just the scaled up variance of the first step. But numerical simulations show that the approximation is not very good, although usable for me at the moment (I only need sets with up to about $20$ positive and negative numbers so far but this may change in the future and the approximation might be even worse then).
Suspicions I have
I have no good approach how to move forward yet I have some suspicion that the binomial distribution with the right parametrization (and some correction term?) could be used. I don't know but perhaps a generating function could be used too but I am not an expert here.
 A: Hint:
Let $X_i, i \in \{a, b, \cdots, z\}$ be independent random variables with values $0$ and $i$, each with probability $\frac{1}{2}$.
Define a new random variable $S = \sum_{i \in \{a, \cdots, z\}} X_i$. Observe that the variance of $\mathbb{S}$ equals to $Var(S)$.

Here is how it works. 
There are $N = 2^n$ combinations in $\mathbb{S}$. For notational convenience, denote these combinations as $\omega_1$, $\omega_2$, $\cdots$, $\omega_N$. We also define a function sum($\omega$) which returns the sum of elements in a combination $\omega$. Then the mean of $\mathbb{S}$, denoted as $\mathbb{E}(\mathbb{S})$, satisfies:
$$
\mathbb{E}(\mathbb{S}) = \frac{1}{2^n}\sum_{i=1}^{2^n}\textbf{sum}(\omega_i) \tag{1}
$$
and the variance of $\mathbb{S}$, denoted as $Var(\mathbb{S})$, satisfies:
$$
Var(\mathbb{S}) = \frac{1}{2^n}\sum_{i=1}^{2^n}(\textbf{sum}(\omega_i) - \mathbb{E}(\mathbb{S}))^2 \tag{2}
$$
We can also consider (1)-(2) from a probabilistic view. Let $S$ be a random variable with values {$\textbf{sum}(\omega_1)$, $\textbf{sum}(\omega_2)$, $\cdots$, $\textbf{sum}(\omega_{2^n})$}, each with probability $\frac{1}{2^n}$. Easy to see that the expectation $\mathbb{E}(S)$ and variance $Var(S)$ satisfy:
$$
\mathbb{E}(S) = \mathbb{E}(\mathbb{S})\text{ and }Var(S) = Var(\mathbb{S})
$$
implying that we only need to consider $\mathbb{E}(S)$ and $Var(S)$ only. The only question is how to generate such an $S$.
The Hint part provides a way to generate such an $S$. It guarantees that
$$
\Pr(S = \textbf{sum}(\omega_i)) = \frac{1}{2^n}, \ 1 \leq i \leq 2^n
$$
Note: we assume $\textbf{sum}(\omega_i) \neq \textbf{sum}(\omega_j),\forall \ i\neq j$ here for convenience.
Now we can proceed to compute $\mathbb{E}(S)$ and $Var(S)$. By linearity of expectation,
$$
\mathbb{E}(S) = \sum_i \mathbb{E}(X_i)  = \frac{1}{2}\sum_{i \in \{a,\cdots, z\}} i
$$
and since $X_i$ are independent,
$$
Var(S) = \sum_i Var(X_i) = \frac{1}{4}\sum_{i \in \{a,\cdots,z\}} i^2
$$
