Sum of Sequence of Cubes/Proof by Induction

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

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

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


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

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.