Evaluate $S=\Sigma\Sigma\Sigma x_ix_jx_k$

Question

I am a novice in this type of sums and I can't even understand the meaning of the three sigmas. Somehow, I am guessing that the answer might be $0$ but I am not sure. I need a well-explained answer(with examples) explaining the meaning and the method to solve it. Lots and lots of thanks beforehand. Sum is written below :

Suppose that $x_1,x_2,...,x_n(n>2)$ are real numbers such that $x_i=-x_{n-i+1}$ for $1\le i\le n$. Consider the sum $S=\Sigma\Sigma\Sigma x_ix_jx_k$, where the summations are taken over all $i,j,k:1\le i,j,k\le n$ and $i,j,k$ are all distinct. Then $S$ equals ____ ?

Answer 1

By problem $S=$ $\displaystyle \sum_{k=1}^n$ $\displaystyle \sum_{j=1}^n$ $\displaystyle \sum_{i=1}^n$ $x_k x_j x_i$. Now notice that,

$$\displaystyle \sum_{k=1}^n \displaystyle \sum_{j=1}^n \displaystyle \sum_{i=1}^n x_k x_j x_i =S=\displaystyle \sum_{k=1}^n \displaystyle \sum_{j=1}^n \displaystyle \sum_{i=1}^n x_k x_j x_{n+i-1}=-\displaystyle \sum_{k=1}^n \displaystyle \sum_{j=1}^n \displaystyle \sum_{i=1}^n x_k x_j x_i$$

The rest part is easy.

Answer 2

The $\sum\sum\sum$ here really means sum over the 3 indices $i,j,k$. However, it is a bad notation and misleading. Nowhere does it capture the requirement $i,j,k$ are distinct. Literally, you should interpret it as

$$\sum\sum\sum\quad\equiv\quad \sum_{i=1}^n\sum_{\substack{j=1\\j\ne i}}^n\sum_{\substack{k=1\\k \ne i,j}}^n$$ A less misleading notation could be something like $\displaystyle\;\sum_{i\ne j\ne k}\;$ instead.

For the question of the sum. Yes, it does vanishes. Notice

$$i,j,k\;\;\text{ all distinct } \quad\iff\quad n-i+1, n-j+1, n-k+1\;\;\text{ all distinct }$$

By a relabeling of indices $x_\alpha \mapsto x_{n-\alpha+1}$, we have

$$\sum_{i\ne j \ne k} x_i x_j x_k = \sum_{i\ne j \ne k} x_{n-i+1} x_{n-j+1} x_{n-k+1}$$

Apply the given condition $x_{n-\alpha+1} = -x_\alpha$, you can rewrite RHS as

$$RHS = \sum_{i\ne j \ne k} (-x_i)(-x_j)(-x_k) = -\sum_{i\ne j \ne k} x_i x_j x_k = -LHS$$

From this, you can conclude $S = LHS = RHS = 0$.