# Square of Triangular Number equals Sum of Sequence of Cubes/Proof 2

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## Theorem

- $\displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$

where $T_n$ denotes the $n$th triangular number.

## Proof

The proof proceeds by induction.

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

- $\displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$

### Basis for the Induction

$\map P 1$ is the case:

\(\ds \sum_{i \mathop = 1}^1 i^3\) | \(=\) | \(\ds 1^3\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {\frac {1 \paren {1 + 1} } 2}^2\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds {T_1}^2\) | Closed Form for Triangular Numbers |

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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

- $\displaystyle \sum_{i \mathop = 1}^k i^3 = {T_k}^2$

from which it is to be shown that:

- $\displaystyle \sum_{i \mathop = 1}^{k + 1} i^3 = {T_{k + 1} }^2$

### Induction Step

This is the induction step:

\(\ds \sum_{i \mathop = 1}^{k + 1} i^3\) | \(=\) | \(\ds \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds {T_k}^2 + \paren {k + 1}^3\) | Induction Hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds {T_k}^2 + \paren { {T_{k + 1} }^2 - {T_k}^2}\) | Cube Number as Difference between Squares of Triangular Numbers | |||||||||||

\(\ds \) | \(=\) | \(\ds {T_{k + 1} }^2\) |

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

Therefore:

- $\forall n \in \Z_{> 0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$

$\blacksquare$

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $15$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $15$