Summation Notation Confusion I am unclear about what the following summation means given that $\lambda_i: \forall i \in \{1,2,\ldots n\}$:
$\mu_{4:4} = \sum\limits_{i=1}^{4} \lambda_i + \mathop{\sum\sum}_{1\leq i_1 < i_2 \leq 4}(\lambda_{i_1} + \lambda_{i_2}) + \mathop{\sum\sum\sum}_{1\leq i_1 < i_2 <i_3 \leq 4}(\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3})$
I understand how this term expands:
$\sum\limits_{i=1}^{4} \lambda_i = \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4$.
But, I don't understand what how this term expands
$\mathop{\sum\sum}_{\substack{1\leq i_1 < i_2 \leq 4}}(\lambda_{i_1} + \lambda_{i_2})$
Nor do I understand how this term expands
$\mathop{\sum\sum\sum}_{\substack{1\leq i_1 < i_2 <i_3 \leq 4}}(\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3})
$
Any help in these matters would be appreciated.
 A: 
I understand how this term expands
$\sum\limits_{i=1}^{4} \lambda_i = \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4$.
But, I don't understand what how this term expands
$\mathop{\sum\sum}_{1\leq i_1 < i_2 \leq 4}(\lambda_{i_1} + \lambda_{i_2})$

The subscript is just another way of indicating the domain of the indices.
Like so: $\displaystyle\quad\quad \sum\limits_{i=1}^{4} \lambda_i = \sum\limits_{1\leq i \leq 4} \lambda_i$
Thus, $1\leq i_1 < i_2 \leq 4$ means: $i_1\in[1\,..\,(i_2-1)], i_2\in[(i_1+1)\,..\,4]$
Hence:
$$\mathop{\sum\sum}_{1\leq i_1 < i_2 \leq 4}(\lambda_{i_1} + \lambda_{i_2}) \\ = \sum\limits_{i_1=1}^{3}\left(\sum\limits_{i_2=i_1+1}^4(\lambda_{i_1} + \lambda_{i_2})\right) \\ = ((\lambda_1+\lambda_2)+(\lambda_1+\lambda_3)+(\lambda_1+\lambda_4))+((\lambda_2+\lambda_3)+(\lambda_2+\lambda_4))+((\lambda_3+\lambda_4)) \\ = 3(\lambda_1 +\lambda_2+\lambda_3+\lambda_4) \\ = 3\sum_{i=1}^4 \lambda_i$$

Nor do I understand how this term expands
$\mathop{\sum\sum\sum}_{1\leq i_1 < i_2 <i_3 \leq 4}(\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3})$

$$\mathop{\sum\sum\sum}_{1\leq i_1 < i_2 <i_3 \leq 4}(\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3}) \\ = \sum_{i_1=1}^2\left(\sum_{i_2=i_1+1}^{3}\left(\sum_{i_3=i_2+1}^{4} (\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3})\right)\right) \\ = (\lambda_1\!+\!\lambda_2\!+\!\lambda_3)\!+\!(\lambda_1\!+\!\lambda_2\!+\!\lambda_4)\!+\!(\lambda_1\!+\!\lambda_3\!+\!\lambda_4)\!+\!(\lambda_2\!+\!\lambda_3\!+\!\lambda_4) \\ = 3( \lambda_1 + \lambda_2+\lambda_3+\lambda_4)\\ = 3\sum_{i=1}^4 \lambda_i$$
A: The $\sum\sum_{1\le i_1\lt i_2\le 4} \ldots $ means that you take the sum over all pairs of $i_1$ and $i_2$ that satisfy the specified condition, namely that $$1\le i_1\lt i_2\le 4.$$  In this case that means that  $i_1$ and $i_2$ should take the following pairs of values: $$\begin{array}{cc}
i_1&i_2\\\hline
1&2\\1&3\\1&4\\2&3\\2&4\\3&4\end{array}$$ so the summation consists of 6 terms.
Similarly the triple sum is the sum over all triples $(i_1, i_2, i_3)$ for which all of  $$1\le i_1\\ i_1\lt i_2\\ i_2\lt i_3\\i_3\le 4$$ all hold, so there are 4 terms in the sum:
$$\begin{array}{ccc}
i_1 & i_2&i_3\\\hline
1 & 2 & 3 \\
1 & 2 & 4 \\
1 & 3 & 4 \\
2 & 3 & 4
\end{array}$$
A: I would interpret ${\sum\sum}_{\substack{1\leq i_1<i_2\leq 4}}(\lambda_{i_1}+\lambda_{i_2})$ as $(\lambda_1+\lambda_2)+(\lambda_1+\lambda_3)+(\lambda_1+\lambda_4)+(\lambda_2+\lambda_3)+(\lambda_2+\lambda_4)+(\lambda_3+\lambda_4)$
I would interpret ${\sum\sum\sum}_{\substack{1\leq i_1 < i_2 <i_3 \leq 4}}(\lambda_{i_1}+\lambda_{i_2}+\lambda_{i_3})$ as $(\lambda_1+\lambda_2+\lambda_3)+(\lambda_1+\lambda_2+\lambda_4)+(\lambda_1+\lambda_3+\lambda_4)+(\lambda_2+\lambda_3+\lambda_4)$
