# Sum of Sequence of Cubes/Proof by Induction

## Theorem

$\displaystyle \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$

## Proof

First, from Closed Form for Triangular Numbers:

$\displaystyle \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

So:

$\displaystyle \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$

Next we use induction on $n$ to show that:

$\displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$

The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$\displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$

### Basis for the Induction

$\map P 1$ is the case:

$1^3 = \dfrac {1 \paren {1 + 1}^2} 4$

 $\displaystyle \sum_{i \mathop = 1}^1 i^3$ $=$ $\displaystyle 1^3$ $\displaystyle$ $=$ $\displaystyle \dfrac {1^2 \paren {1 + 1}^2} 4$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

$\displaystyle \sum_{i \mathop = 1}^k i^3 = \dfrac {k^2 \paren {k + 1}^2} 4$

from which it is to be shown that:

$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3 = \dfrac {\paren {k + 1}^2 \paren {k + 2}^2} 4$

### Induction Step

This is the induction step:

 $\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3$ $=$ $\displaystyle \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3$ $\displaystyle$ $=$ $\displaystyle \frac {k^2 \paren {k + 1}^2} 4 + \paren {k + 1}^3$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \frac {k^4 + 2 k^3 + k^2} 4 + \frac {4 k^3 + 12 k^2 + 12 k + 4} 4$ $\displaystyle$ $=$ $\displaystyle \frac {k^4 + 6 k^3 + 13 k^2 + 12 k + 4} 4$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {k + 1}^2 \paren {k + 2}^2} 4$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\forall n \in \Z_{>0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$