2026/015) Show that difference between squares of 2 conscutive triangular numbers is a perfect cube

We have $T_n = frac{n(n+1)}{2}$ 

To avoid fraction we have $2T_n = n(n+1)$

Now we have 

$ (2T_{n+1})^2 -   (2T_{n})^2 = ((n+1)(n+2))^2 -((n+1)(n+2))^2 $

Or $4(T_{n+1}^2-T_n^2) = (n+1)^2((n+2)^2 - n^2) = (n+1)^2 * 4 (n+1))= 4(n+1)^3$

Or   $T_{n+1}^2 -   T_{n}^2 = (n+1)^3$