Sum of Sequence of Cubes

Theorem

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

Proof 1

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

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


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

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

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

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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:

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Hence the result.

Direct Proof by Recursion
Let $$A(n) = 1 + 2 + \cdots + n = \sum_{i=1}^{n}i = \frac{n(n+1)}{2}$$.

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

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

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