How do you solve the expression
$$\sum_{j=0}^nj\sum_{1\le i_1<i_2<...<i_j\le n}m_{i_1}+m_{i_2}+...+m_{i_j}$$
with the following:
$$ \begin{array}{ll} variable & value\\ \hline n & 3\\ m_1 & 1 \\ m_2 & 2 \\ m_3 & 3 \\ \end{array} $$
I get that $$ 0 * \text{something}\\ + 1 * \text{something}\\ + 2 * \text{something}\\ + 3 * \text{something} $$
?
But I can't figure out how to express that $\text{something}$ from the original sigma notation.
On
when $j=1$ we have $$ \sum_{1\le i_1<i_2<...<i_j\le n}m_{i_1}+m_{i_2}+...+m_{i_j} = \sum_{1 \le i_1 \le 3} m_{i_1} = m_1+m_2+m_3 $$ When $j=2$ we have $$ \sum_{1\le i_1<i_2<...<i_j\le n}m_{i_1}+m_{i_2}+...+m_{i_j} = \sum_{1\le i_1 < i_2 \le 3} (m_{i_1}+m_{i_2}) = (m_1+m_2)+(m_1+m_3)+(m_2+m_3) $$ For $j=3$ we have $$ \sum_{1\le i_1<i_2<...<i_j\le n}m_{i_1}+m_{i_2}+...+m_{i_j} =\sum_{1 \le i_1 < i_2 < i_3 \le 3} (m_{i_1}+m_{i_2}+m_{i_3}) =(m_1+m_2+m_3) $$
On
The general formula is: $$S=\sum_{j=1}^{n} j\cdot C_{j-1}^{n-1} \cdot \sum_{k=1}^{n} m_k.$$ Note when $n=4$: $$\begin{align} & j=1; 1\le i_1\le 4; (1),(2),(3),(4); 1(m_1+m_2+m_3+m_4);\\\\ &j=2; 1\le i_1<i_2\le 4; (1,2),(1,3),(1,4),(2,3),(2,4),(3,4); \\3(m_1+m_2+m_3+m_4); \\\\ & j=3; 1\le i_1<i_2<i_3\le 4; (1,2,3),(1,2,4),(1,3,4),(2,3,4); \\\\3(m_1+m_2+m_3+m_4); & j=4; 1\le i_1<i_2<i_3<i_4\le 4; (1,2,3,4); 1(m_1+m_2+m_3+m_4)\end{align}.$$
Comment:
In (1) we write for each $j=1,2,3$ the inner sum, skipping the sum with $j=0$ which does not contribute anything.
In (2) we explicitly write the terms of each sum which is feasible since $n=3$ is small. Observe that in (2) the first term $m_1+m_2+m_3$ are three summands, whereas the third term $3(m_1+m_2+m_3)$ is one summand.
Note: It is crucial to use brackets in the first line, otherwise the meaning is \begin{align*} \sum_{1\le i_1<i_2<...<i_j\le 3}&m_{i_1}+m_{i_2}+\cdots+m_{i_j}\\ &=\left(\sum_{1\le i_1<i_2<...<i_j\le 3}m_{i_1}\right)+m_{i_2}+\cdots+m_{i_j}\\ &=m_{i_2}+\cdots +m_{i_j}+\sum_{1\le i_1<i_2<...<i_j\le 3}m_{i_1} \end{align*} which is usually not the intention.