Sum of Sequence of Cubes/Proof by Recursion

Theorem

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

Proof

Let $\displaystyle A \left({n}\right) = 1 + 2 + \cdots + n = \sum_{i=1}^n i = \frac{n \left({n + 1}\right)} 2$.

Let $\displaystyle B \left({n}\right) = 1^2 + 2^2 + \cdots + n^2 = \sum_{i=1}^n i^2 = \frac{n \left({n + 1}\right) \left({2 n + 1}\right)} 6$.

Let $\displaystyle S \left({n}\right) = 1^3 + 2^3 + \cdots + n^3 = \sum_{i=1}^n i^3$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \left({n}\right)\)	\(=\)	\(\displaystyle n^3 + \left({n - 1}\right)^3 + \left({n - 2}\right)^3 + \cdots + 2^3 + 1^3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=0}^{n-1} \left({n - k}\right)^3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=0}^{n-1} n^3 - 3 n^2 k + 3 n k^2 - k^3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n^4 - 3 n^2 \cdot A \left({n - 1}\right) + 3 n \cdot B \left({n - 1}\right) - S \left({n - 1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \left({n}\right)\)	\(=\)	\(\displaystyle n^4 - 3n^2 \cdot \frac{n \left({n - 1}\right)} 2 + 3n \cdot \frac{n \left({n - 1}\right) \left({2 n - 1}\right)} 6 - S \left({n - 1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	alternate recursive definition
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \left({n}\right)\)	\(=\)	\(\displaystyle n^3 + S \left({n - 1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	standard recursive definition of the sum of sequence of cubes
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 2 S \left({n}\right)\)	\(=\)	\(\displaystyle n^4 + n^3 - 3n^2 \cdot \frac{n \left({n - 1}\right)} 2 + 3n \cdot \frac{n \left({n - 1}\right) \left({2 n - 1}\right)} 6\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	so we add the eqns together ...
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{n^2 \left({n + 1}\right)^2} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	... and simplify/factor the result
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \left({n}\right)\)	\(=\)	\(\displaystyle \frac{n^2 \left({n + 1}\right)^2} 4\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$