Sum of Sequence of Cubes/Proof by Recursion

Theorem

 * $\displaystyle \sum_{i \mathop = 1}^n i^3 = \left({\sum_{i \mathop = 1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4$

Proof
From Closed Form for Triangular Numbers‎:
 * $(1): \quad \displaystyle A \left({n}\right) := \sum_{i \mathop = 1}^n i = \frac{n \left({n + 1}\right)} 2$

From Sum of Sequence of Squares:
 * $(2): \quad \displaystyle B \left({n}\right) := \sum_{i \mathop = 1}^n i^2 = \frac{n \left({n + 1}\right) \left({2 n + 1}\right)} 6$

Let $\displaystyle S \left({n}\right) = \sum_{i \mathop = 1}^n i^3$.

Then: