# How do I read this triple summation? $\sum_{1\leq i < j < k \leq 4}a_{ijk}$

How do I read this triple summation?

$$\sum_{1\leq i < j < k \leq 4}a_{ijk}$$

The exercise asks me to express it as three sumations and to expand them in the following way:

1) Summing first on $k$, then on $j$ and last on $i$.

2) Summing first on $i$, then on $j$ and last on $k$.

My attempt:

$$\sum_{k=3}^4\sum_{j=2}^{k-1}\sum_{i=1}^{j-1} a_{ijk}=\sum_{j=2}^2\sum_{i=1}^{j-1}a_{ij3}+\sum_{j=2}^3\sum_{i=1}^{j-1}a_{ij4}=\sum_{i=1}^1a_{i23}+\sum_{i=1}^{1}a_{i24}+\sum_{i=1}^2a_{i34} \\=a_{123}+a_{124}+a_{134}+a_{234}$$

Is this correct so far? I don't know how to do the following parts of the exercise.

Could someone give me a general explanation on how to read this type of multiple summations?

$$\sum_{i=1}^2 \sum_{j=i+1}^3 \sum_{k=j+1}^4 a_{ijk} = \sum_{k=3}^4 \sum_{j=2}^{k-1} \sum_{i=1}^{j-1} a_{ijk} = a_{123}+a_{124}+a_{134}+a_{234}$$

• Could you explain how you went the one starting with k to the other one? May 20 '15 at 20:57
• I didn't go from one to the other; they're constructed essentially identically, but the left triple sum is constructed from the top, and the right triple sum is constructed from the bottom. For instance, if $1 \leq i < j < k \leq 4$, then $k$ can only run from $3$ to $4$ (if $k \leq 2$ there isn't enough "room" for $i$ and $j$), then $j$ can only run from $2$ to $k-1$, and $i$ can only run from $1$ to $j-1$. May 20 '15 at 22:14

You are correct.

For (1), $\displaystyle\sum_{1\leq i<j<k\leq 4}a_{ijk}=\sum_{1\leq i<j< 3}a_{ij3}+\sum_{1\leq i<j<4}a_{ij4}=\sum_{1\leq i<2}a_{i23}+\sum_{1\leq i<2}a_{i24}+\sum_{1\leq i<3}a_{i34}=a_{123}+a_{124}+a_{134}+a_{234}.$

(2) is analogous.

1) $\sum_{k=1}^4 \sum_{j=1}^{k-1} \sum_{i=1}^{j-1} a_{ijk}$

2) $\sum_{i=1}^4 \sum_{j=i+1}^4 \sum_{k=j+1}^4 a_{ijk}$

When there is just one part of the sum, you should think that you're summing over a set. You should think of it as

$$\sum_{(i,j,k)\in S}a_{ijk}$$

where $S=\{(i,j,k)\mid1\le i<j<k\le4\}$