I know that the expansion of the square of a summation can be expressed as:
$$\left( \sum_{n=1}^N a_n \right)^2 = \sum_{n=1}^N a_n^2 + 2 \sum_{j=1}^{N}\sum_{i=1}^{j-1} a_i a_j$$
where $a_n \in \mathbb{R}\backslash\{0\}$ or $a_n \in \mathbb{Z}\backslash\{0\}$ for $n=1,…,N$ (see for example this).
This seems, of course, just a neat elegant formulate to facilitate the usage of the multinomial theorem for when exponent equals to $2$. I wonder if there is a similarly neat elegant formula for when the exponent is equal to $3$. Or, in other words, an analogous formula to expand $\left( \sum_{n=1}^N a_n \right)^3$ that saves the time of having to resort to the multinomial theorem itself.
In a previous question in this website it is said that it would be something like:
$$\sum_{n=1}^N a_n^3+3\sum_{i\neq j}^Na_i^2 a_j+6\sum_{i\neq j\neq k}^Na_ia_ja_k$$
However, I agree with a comment there that has never been replied to. Shouldn't it be instead:
$$\sum_{n=1}^N a_n^3+3\sum_{i<j}^Na_i^2 a_j+6\sum_{i< j< k}^Na_ia_ja_k$$
Best Answer
$$\left( \sum_{n=1}^N a_n \right)^3 = \sum_{i=1}^N a_i^3 + 3 \sum_{i=1}^{N}\sum_{j=1}^{i-1} a_i^2 a_j+ 3 \sum_{i=1}^{N}\sum_{j=1}^{i-1} a_i a_j^2+6 \sum_{i=1}^{N}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1} a_i a_j a_k$$