Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3/Proof 2

Proof
We have that:


 * $\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}$

From Summation of Products of n Numbers taken m at a time with Repetitions:


 * $\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac { {S_2}^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac { {S_m}^{k_m} } {m^{k_m} k_m !}$

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

Setting $m = 3$:

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 $\tuple {k_1, k_2, k_3}$ can be:


 * $\tuple {3, 0, 0}$
 * $\tuple {1, 1, 0}$
 * $\tuple {0, 0, 1}$

Hence: