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

 * 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 $$n+1\,$$.

$$ $$ $$ $$ $$

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