Newton-Girard Formulas/Examples/Order 3

Theorem
Let $a, b \in \Z$ be integers such that $b \ge a$.

Let $U$ be a set of $n = b - a + 1$ numbers $\left\{ {x_a, x_{a + 1}, \ldots, x_b}\right\}$.

Then:
 * $\displaystyle \sum_{a \mathop \le i \mathop < j \mathop < k \mathop \le b} x_i x_j x_k = \dfrac { {S_1}^3} 6 - \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$

where:
 * $\displaystyle S_r := \sum_{k \mathop = a}^b {x_k}^r$

Proof
From Newton-Girard Formulas:


 * $\displaystyle \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\left({-S_2}\right)^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \cdots \dfrac {\left({\left({-1}\right)^{m + 1} S_m}\right)^{k_m} } {m^{k_m} k_m !}$

where:
 * $S_r = \displaystyle \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$.

Setting $m = 3$, and setting $x_i := x_{j_1}, x_j := x_{j_2}, x_k := x_{j_3}$:

We need to find all sets of $k_1, k_2, k_3 \in \Z_{\ge 0}$ such that:


 * $k_1 + 2 k_2 + 3 k_3 = 3$

Thus $\left({k_1, k_2, k_3}\right)$ can be:


 * $\left({3, 0, 0}\right)$
 * $\left({1, 1, 0}\right)$
 * $\left({0, 0, 1}\right)$

Hence: