Cube Number as Difference between Squares of Triangular Numbers/Proof 2

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.

$\ds n^3$

$=$

$\ds \sum_{k \mathop = 1}^n k^3 - \sum_{k \mathop = 1}^{n - 1} k^3$

$\ds $

$=$

$\ds \paren {\frac {n^2 \paren {n + 1}^2} 4} - \paren {\frac {\paren {n - 1}^2 n^2} 4}$

$\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 $

$=$

$\ds {T_n}^2 - {T_{n - 1} }^2$

$\blacksquare$