1

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.

useranonis
  • 237

3 Answers3

0

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 $$ \sum_{1\le i_1<i_2<...<i_j\le n}m_{i_1}+m_{i_2}+...+m_{i_j} =\sum_{1 \le i_1 < i_2 < i_3 \le 3} (m_{i_1}+m_{i_2}+m_{i_3}) =(m_1+m_2+m_3) $$

GEdgar
  • 111,679
0

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 $n=4$: $$\begin{align} & j=1; 1\le i_1\le 4; (1),(2),(3),(4); 1(m_1+m_2+m_3+m_4);\\\\ &j=2; 1\le i_1<i_2\le 4; (1,2),(1,3),(1,4),(2,3),(2,4),(3,4); \\3(m_1+m_2+m_3+m_4); \\\\ & j=3; 1\le i_1<i_2<i_3\le 4; (1,2,3),(1,2,4),(1,3,4),(2,3,4); \\\\3(m_1+m_2+m_3+m_4); & j=4; 1\le i_1<i_2<i_3<i_4\le 4; (1,2,3,4); 1(m_1+m_2+m_3+m_4)\end{align}.$$

useranonis
  • 237
farruhota
  • 31,482
0

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.

Markus Scheuer
  • 108,315