Square of Triangular Number equals Sum of Sequence of Cubes

Theorem

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

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

Proof 1

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

\(=\)

\(\ds \frac {n^2 \paren {n + 1}^2} 4\)

Sum of Sequence of Cubes

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds {T_n}^2\)

Closed Form for Triangular Numbers

$\blacksquare$

Proof 2

The proof proceeds by induction.

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

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

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:

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

from which it is to be shown that:

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

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:

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

$\blacksquare$