Sum of Sequence of Cubes

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 1

 * First, from Closed Form for Triangular Numbers, we have that:
 * $\displaystyle \sum_{i=1}^n i = \frac {n \left({n + 1}\right)} 2$

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


 * Next we use induction on $n$ to show that $\displaystyle \sum_{i=1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4$.

The base case holds since $\displaystyle 1^3 = \frac{1 \left({1 + 1}\right)^2} 4$.

Now we need to show that if it holds for $n$, then it holds for $n + 1$.

By the Principle of Mathematical Induction, the proof is complete.

Proof 2
By Nicomachus's Theorem, we have:


 * $\forall n \in \N^*: n^3 = \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$

Also by Nicomachus's Theorem, we have that the first term for $\left({n + 1}\right)^3$ is $2$ greater than the last term for $n^3$.

So if we add them all up together, we get:

Hence the result.

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

Historical Note
This result was documented by Āryabhaṭa in his work Āryabhaṭīya of 499 CE.