Cube Number as Difference between Squares of Triangular Numbers/Proof 1

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

\(\ds {T_n}^2 - {T_{n - 1} }^2\)

\(=\)

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

Closed Form for Triangular Numbers

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \frac {\paren {n^4 + 2 n^3 + n^2} - \paren {n^4 - 2 n^3 + n^2} } 4\)

\(\ds \)

\(=\)

\(\ds \frac {4 n^3} 4\)

\(\ds \)

\(=\)

\(\ds n^3\)

$\blacksquare$