How to re-express sigma notation with sub index 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.
 A: 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
$$
\sum_{1\le i_1<i_2<...<i_j\le n}m_{i_1}+m_{i_2}+...+m_{i_j}
=\sum_{1 \le i_1 < i_2 < i_3 \le 3} (m_{i_1}+m_{i_2}+m_{i_3})
=(m_1+m_2+m_3)
$$
A: 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 $n=4$:
$$\begin{align} & j=1; 1\le i_1\le 4; (1),(2),(3),(4); 1(m_1+m_2+m_3+m_4);\\\\ 
&j=2; 1\le i_1<i_2\le 4; (1,2),(1,3),(1,4),(2,3),(2,4),(3,4); \\3(m_1+m_2+m_3+m_4); \\\\
& j=3; 1\le i_1<i_2<i_3\le 4; (1,2,3),(1,2,4),(1,3,4),(2,3,4); \\\\3(m_1+m_2+m_3+m_4);
& j=4; 1\le i_1<i_2<i_3<i_4\le 4; (1,2,3,4); 1(m_1+m_2+m_3+m_4)\end{align}.$$
A: 
We obtain for $n=3$ and $m_i=i, 1\leq i\leq 3$:
\begin{align*}
\color{blue}{\sum_{j=0}^3}&\color{blue}{j\sum_{1\le i_1<i_2<...<i_j\le 3}\left(m_{i_1}+m_{i_2}+...+m_{i_j}\right)}\\
&=\sum_{1\leq i_1\leq 3}m_{i_1}
+2\sum_{1\leq i_1< i_2\leq 3}\left(m_{i_1}+m_{i_2}\right)+3\sum_{1\leq i_1<i_2<i_3\leq 3}\left(m_{i_1}+m_{i_2}+m_{i_3}\right)\tag{1}\\
&=\left(m_1+m_2+m_3\right)+2\left[(m_1+m_2)+(m_1+m_3)+(m_2+m_3)\right]\\
&\qquad +3(m_1+m_2+m_3)\tag{2}\\
&=8(m_1+m_2+m_3)\\
&=8 (1+2+3)\\
&\color{blue}{\,=48}
\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.
