Cube Number as Difference between Squares of Triangular Numbers

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Theorem

Let $n \in \Z_{>0}$ be a positive integer.

Then:

$n^3 = {T_n}^2 - {T_{n - 1} }^2$

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


Proof 1

\(\ds {T_n}^2 - {T_{n - 1} }^2\) \(=\) \(\ds \paren {\frac {n \paren {n + 1} } 2}^2 - \paren {\frac {\paren {n - 1} n} 2}^2\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds \frac {\paren {n^2 + n}^2 - \paren {n^2 - n}^2} 4\)
\(\ds \) \(=\) \(\ds \frac {\paren {n^4 + 2 n^3 + n^2} - \paren {n^4 - 2 n^3 + n^2} } 4\)
\(\ds \) \(=\) \(\ds \frac {4 n^3} 4\)
\(\ds \) \(=\) \(\ds n^3\)

$\blacksquare$


Proof 2

\(\ds n^3\) \(=\) \(\ds \sum_{k \mathop = 1}^n k^3 - \sum_{k \mathop = 1}^{n - 1} k^3\)
\(\ds \) \(=\) \(\ds \paren {\frac {n^2 \paren {n + 1}^2} 4} - \paren {\frac {\paren {n - 1}^2 n^2} 4}\) Sum of Sequence of Cubes
\(\ds \) \(=\) \(\ds \frac {n^2 \paren {\paren {n + 1}^2 - \paren {n - 1}^2} } 4\)
\(\ds \) \(=\) \(\ds \paren {\frac {n \paren {n + 1} } 2}^2 - \paren {\frac {n \paren {n - 1} } 2}^2\)
\(\ds \) \(=\) \(\ds {T_n}^2 - {T_{n - 1} }^2\) Closed Form for Triangular Numbers

$\blacksquare$


Sources

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