Double index summation: $\sum_{ 1 \leq i < j \leq 3 }(2i+j). $ I understand multiple summations, but I do not know how to do this summation. Could someone break this down for me?
For example:
 $$\sum_{ 1 \leq i < j \leq 3 }(2i+j). $$
 A: With two parameters $i$ and $j$ under the sum, this is a double sum. It means you have to find all $i$ and $j$ that are $1$, $2$, or $3$, and also $i < j$. Thus, the possible pairs are $$(i, j) \in \{(1, 2), (1, 3), (2, 3)\}.$$
Then, you plug in each pair into the expression $2i + j$ and add it all up: 
$$[2(1) + 2] + [2(1) + 3] + [2(2) + 3] = 4 + 5 + 7 = 16.$$
A: This notation is telling you to take the sum over all terms $2i + j$ where $1 \leq i \lt j \leq 3$. That being said, let's just write it out.
$$
\sum_{1 \leq i \lt j \leq 3}2i + j = (2\cdot 1 + 2) + (2\cdot 1 + 3) + (2\cdot 2 + 3).
$$
In this case, there's only three terms to sum. Note that we take the sum over all ordered pairs $(i, j)$ where $1 \leq i \lt j \leq 3$.
A: As has been stated,
this is a double sum.
If we let the
upper limit be $n$
instead of $3$,
it can be explicitly converted
into a double sum
and then evaluated like this:
$\begin{array}\\
\sum_{ 1 \leq i < j \leq n }(2i+j)
&=\sum_{1 < j \le n}\sum_{1\le i < j}(2i+j)\\
&=\sum_{j=2}^{ n}\sum_{i=1}^{j-1}(2i+j)\\
&=\sum_{j=2}^{ n}\sum_{i=1}^{j-1}(2i)+\sum_{j=2}^{n}\sum_{i=1}^{j-1}(j)\\
&=\sum_{j=2}^{ n}2\dfrac{(j-1)j}{2}+\sum_{j=2}^{n}j(j-1)\\
&=2\sum_{j=2}^{n}j(j-1)\\
\end{array}
$
and this you should be
able to evaluate.
A: I think it is clearer to call $a_{ij} = 2i+j$ to explain this. We must sum over all pairs $(i,j)$ satisfying $1\leq i < j \leq 3$, that is: $$\sum_{ 1\leq i < j \leq 3 } a_{ij} = a_{12} + a_{13} + a_{23} = 4 + 5 + 7 = 16. $$
In general you can rewrite it as a double summation: $$\sum_{1 \leq i < j\leq n} a_{ij} = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_{ij}.$$
