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

Theorem

 * $\displaystyle \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^i \sum_{k \mathop = 0}^j a_i a_j a_k = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$

where:
 * $\displaystyle S_r := \sum_{i \mathop = 0}^n {a_i}^r$

Proof
Let:

Also, let:

Hence:

Let:

as calculated above.

Thus:
 * $(1): \quad 2 A = A_1 + A_3$

Similarly:

Then:

Let:

as calculated above.

Thus:
 * $(2): \quad A_1 + A = A_2 + A_4$

Then:

Now we have that:

and so:

Finally: