Expansion of the square of the sum of $N$ numbers Do I need to cite any results to use the following equality
$$\left( \sum_{n=1}^N a_n \right)^2 = \sum_{n=1}^N a_n^2 + 2 \sum_{j=1}^{N}\sum_{i=1}^{j-1} a_i a_j $$
where $a_n \in \mathbb{R}\setminus{}\left\{0\right\} $ or $a_n \in \mathbb{Z}\setminus{}\left\{0\right\} $ for $n = 1,\ldots{},N$? It is the definition of the second exponent of sum of $N$ numbers.
I believe I have found a result related to the above expansion. It looks like it is a special case of the following equation (let $x=1$ and $a_0 = a_i = 0$ for $i>N$ ) which is given in (8) in the Power Sum page of WolframMath:
$$\left( \sum_{n=0}^\infty a_n x^n \right)^2 = \sum_{n=0}^\infty a_n^2 x^{2n} + 2 \sum_{\substack{n=1\\i+j=n\\i<j}}^{\infty} a_i a_j x^n.$$
However, I could not find any citations to the reference of this result.
 A: There is a simple direct proof.
$$
\left(\sum_{n=1}^N a_n \right)^2 = \left(\sum_{n=1}^N a_n \right)\left(\sum_{m=1}^N a_m \right)
= \sum_{n=1}^N \sum_{m=1}^N a_na_m =
\sum_{n=1}^N a_n^2 + \sum_{n=1}^N \sum_{m=1\atop m\ne n}^N a_na_m,
$$
and for the last sum we have
$$
\sum_{n=1}^N \sum_{m=1\atop m\ne n}^N a_na_m = 2 \sum_{n=1}^N \sum_{m=1\atop m < n}^N a_na_m
= 2 \sum_{n=1}^N \sum_{m=1}^{n-1} a_na_m.
$$
A: I think this is well-known, but whether this can be used without proof depends on in which level of mathematics you are studying in. In fact there is an easy proof by induction on $N$.
When $N = 1$, then L.H.S. $ = a_1^2 = $ R.H.S.
Assume
$$\left(\sum_{n=1}^N a_n\right)^2 = \sum_{n=1}^N a_n^2 + 2 \sum_{j=1}^N \sum_{i=1}^{j-1} a_ia_j$$
Then
\begin{align}
\left(\sum_{n=1}^{N+1} a_n\right)^2
& = a_{N+1}^2 + 2a_{N+1}\sum_{n=1}^N a_n + \left(\sum_{n=1}^N a_n\right)^2\\
& = a_{N+1}^2 + 2a_{N+1}\sum_{i=1}^N a_i + \left(\sum_{n=1}^N a_n^2 + 2 \sum_{j=1}^N \sum_{i=1}^{j-1} a_ia_j\right)\\
& = \left(\sum_{n=1}^N a_n^2 + a_{N+1}^2\right) + 2\left(\sum_{j=1}^N \sum_{i=1}^{j-1} a_ia_j + \sum_{i=1}^{(N+1)-1} a_ia_{N+1}\right)\\
& = \sum_{n=1}^{N+1} a_n^2 + 2\sum_{j=1}^{N+1} \sum_{i=1}^{j-1} a_ia_j
\end{align}
