The answer is \(2^7 \cdot 552\). We prove the more general formula \(2^{n-3}(5n^2 + 5n + 2)\), replacing the \(10\) with \(n\geq 3\). Let \(f(X)\) denote the sum of all elements of the set \(X\), and let \(g(X)\) denote the sum \(a_1 + a_3 + a_5 + \cdots\), where \(X=\{a_1,a_2,a_3,\ldots\}\) is arranged in increasing order. Then what we want is \[2\left(\sum_{X\subseteq [n]} f(X)\right) + \left(\sum_{X\subseteq [n]} g(X)\right)\] First we calculate the first sum. To do this, note that in the sum, each number \(a\in[n]\) is added exactly \(2^{n-1}\) times; it is added once for each subset, and there are \(2^{n-1}\) such subsets (\(2\) options for each of the other \(n-1\) numbers). Thus, this sum evaluates to \[\sum_{a=1}^n 2^{n-1}\cdot a = n(n+1)\cdot 2^{n-2}\]and multiplying by \(2\) gives us the first part, which is \(n(n+1)\cdot 2^{n-1}\). Now for the second sum (which is the meat of this solution): \[\sum_{X\subseteq [n]} g(X)\] To evaluate this, we again count how many times each \(a\in[n]\) is summed. Suppose \(a=1\). Then, obviously, it is added \(2^{n-1}\) times in the sum. Now assume that \(a\geq 2\). The number of subsets such that \(a\) is at the \((2k+1)\)-th position is \[2^{n-a}\cdot \binom{a-1}{2k}\]where \(2^{n-a}\) is the number of ways to add elements greater than \(a\) to the subset, and \(\binom{a-1}{2k}\) is the number of ways to choose the first \(2k\) elements of the set. Thus, the second sum is \[2^{n-1} + \sum_{a=2}^{n} \sum_{k=0}^{\infty} a\binom{a-1}{2k}\cdot 2^{n-a}\](this is obviously finite assuming \(\binom{n}{k} = 0\) for \(n<k\)) We can sum this as \begin{align*} \sum_{a=2}^{n} \sum_{k=0}^{\infty} a\binom{a-1}{2k}\cdot 2^{n-a} &= \sum_{a=2}^{n} a2^{n-a} \left(\sum_{k=0}^{\infty} \binom{a-1}{2k}\right) \\ &= \sum_{a=2}^{n} a\cdot 2^{n-a} \cdot 2^{a-2} \\ &= 2^{n-2} \sum_{a=2}^n a \\ &= n(n+1)2^{n-3} - 2^{n-2} \\ \end{align*} and adding the \(2^{n-1}\) from before (the case \(a=1\)) gives us the expression \[n(n+1)2^{n-3} + 2^{n-2}\] Thus, our answer must be \[n(n+1)2^{n-1} + n(n+1)2^{n-3} + 2^{n-2}\]which does indeed equal \[2^{n-3} (5n^2 + 5n + 2)\]finishing the proof. Now we can easily plug in \(n=10\) to find the answer to the original question.

Let's solve a more general form, where $S=\{1,2,...,n\}$. For some functions $f$ and $g$, define the WUMBO of a subset $A$ as $f(a_1)+g(a_2)+f(a_3)+g(a_4)+\cdots$. We can get the WSUM of the subset $A$ by setting $f(m)=3m$ and $g(m)=2m$. Now, let $T$ be the sum of all the WUMBOs over all subsets of $S$. Consider one element of the set, let's say $m>1$. The number of ways for $m$ to appear as the oddth term of a subset $A$ is equal to number of subsets from $\{1,2,...,m-1\}$ which has an even number of elements, times the number of subsets $\{m+1,m+2,\ldots,n\}$ has. This is just equal to $2^{m-2}\cdot2^{n-m}=2^{n-2}$. This means, the number of times the term $f(m)$ appears in $T$ is equal to $2^{n-2}$. The number of ways for $m$ to appear as the eventh term of a subset $A$ is equal to number of subsets from $\{1,2,...,m-1\}$ which has an odd number of elements, times the number of subsets $\{m+1,m+2,\ldots,n\}$ has. This is just equal to $2^{m-2}\cdot2^{n-m}=2^{n-2}$ as well. This means, the number of times the term $g(m)$ appears in $T$ is equal to $2^{n-2}$. Now, $1$ is an exception, as it always appears as the first term of a subset $A$. This means, the term $f(1)$ appears $2^{n-1}$ times in $T$, but $g(1)$ never appears. Therefore, the sum of the WUMBOs of all the subsets of $S$ is given by: $T=2^{n-1}f(1)+\sum_{m=2}^n2^{n-2}(f(m)+g(m))$ And finally, since we're finding the sum of WSUMs over all subsets of $S$, we set $f(m)=3m$ and $g(m)=2m$ to get: $2^{n-1}\cdot3\cdot1+\sum_{m=2}^n2^{n-2}(3m+2m)=2^{n-1}\cdot3+2^{n-2}\sum_{m=2}^n5m=\boxed{2^{n-3}(5n^2+5n+2)}$ which is the same formula as Aops29 had.