# Squared dobule sum expression?

Is there a way to get expression for squared double sum?

$$\left(\sum\limits_{i=1}^n \sum\limits_{j=1}^n a_i a_j\right)^2 = \left(\sum\limits_{i=1}^n \sum\limits_{j=1}^n a_i a_j\right)\left(\sum\limits_{k=1}^n \sum\limits_{l=1}^n a_k a_l\right) = ?$$

Something like expression for single squared sum, given here What is the square of summation?.

Edit:

I managed to get next formula by guessing (looking expression for $n=2$):

\begin{align} \left(\sum\limits_{i=1}^n \sum\limits_{j=1}^n a_i a_j\right)^2 &= \sum\limits_{i=1}^n \sum\limits_{j=1}^n a_i^2 a_j^2 + \sum\limits_{i=1}^n \sum\limits_{j \neq l}^n a_i^2 a_j a_l + \sum\limits_{i \neq k}^n \sum\limits_{j=1}^n a_i a_k a_j^2 + \sum\limits_{i \neq k}^n \sum\limits_{j \neq l}^n a_i a_k a_j a_l \\ &= \sum\limits_{i=1}^n \sum\limits_{j=1}^n a_i^2 a_j^2 + 2\sum\limits_{i \neq k}^n \sum\limits_{j=1}^n a_i a_k a_j^2 + \sum\limits_{i \neq k}^n \sum\limits_{j \neq l}^n a_i a_k a_j a_l \end{align}

but I don't know is it correct or how to prove it.

The formula is correct.

\begin{align} \left(\sum_{i=1}^n \sum_{j=1}^n a_i a_j\right)^2 &= \left(\sum\limits_{i=1}^n \sum\limits_{j=1}^n a_i a_j\right)\left(\sum\limits_{k=1}^n \sum\limits_{l=1}^n a_k a_l\right) \\ &= \sum_{i=1}^n \sum_{k=1}^n \sum_{j=1}^n \sum_{l=1}^n a_i a_k a_j a_l. \end{align}

Note that in addition to distributing the $a_i a_j$ terms over the last two summation symbols, we change the order of summation (or more simply we can just swap the symbols $j$ and $k$).

Now let's add up the terms for which $i=k$ separately from the other terms (that is, the terms for which $i\neq k$):

$$\sum_{i=1}^n \sum_{k=1}^n \sum_{j=1}^n \sum_{l=1}^n a_i a_k a_j a_l = \sum_{i=1}^n \sum_{j=1}^n \sum_{l=1}^n a_i^2 a_j a_l + \sum_{\substack{i\neq k \\ 1\leq i\leq n \\ 1\leq k\leq n}} \sum_{j=1}^n \sum_{l=1}^n a_i a_k a_j a_l.$$

Next, within each of the two summations on the right-hand side of the equation above, add up the terms for which $j = l$ separately from the terms for which $j \neq l$:

\begin{align} \sum_{i=1}^n \sum_{j=1}^n \sum_{l=1}^n a_i^2 a_j a_l &= \sum_{i=1}^n \sum_{j=1}^n a_i^2 a_j^2 + \sum_{i=1}^n \sum_{\substack{j\neq l \\ 1\leq j\leq n \\ 1\leq l\leq n}} a_i^2 a_j a_l. \\ \sum_{\substack{i\neq k \\ 1\leq i\leq n \\ 1\leq k\leq n}} \sum_{j=1}^n \sum_{l=1}^n a_i a_k a_j a_l &= \sum_{\substack{i\neq k \\ 1\leq i\leq n \\ 1\leq k\leq n}} \sum_{j=1}^n a_i a_k a_j^2 + \sum_{\substack{i\neq k \\ 1\leq i\leq n \\ 1\leq k\leq n}} \sum_{\substack{j\neq l \\ 1\leq j\leq n \\ 1\leq l\leq n}} a_i a_k a_j a_l. \end{align}

Add these four parts of the sum all together, accounting for the fact that $$\sum_{i=1}^n \sum_{\substack{j\neq l \\ 1\leq j\leq n \\ 1\leq l\leq n}} a_i^2 a_j a_l = \sum_{\substack{i\neq k \\ 1\leq i\leq n \\ 1\leq k\leq n}} \sum_{j=1}^n a_i a_k a_j^2,$$ and you have your formula.