# Thread: Different of two consecutive triangulars is always a cube

1. ## Different of two consecutive triangulars is always a cube

Show that the difference between two consecutive triangular numbres is always a cube.

Proof. Let x be a triangular number, then x = k(k+1)/2 for some integers k.
Let y be a consecutive triangular number, then y = (k+1)(k+2)/2.

y-x = (k+2-k)(k+1)/2 = k+1

But then there isn't any promise that it is a cube?

2. You have stated the theorem incorrectly. The correct statement is:

$\displaystyle \color{white}.\quad.$ “The difference between the squares of two consecutive triangular numbers is always a cube.”

Now try proving that.

3. thanks, I solved the problem now!

$\displaystyle \begin{array}{ccccccccccccccc} \Delta_n & 1 & & 3 & & 6 & & 10 & & 15 & & 21 & & \hdots\\ \text{diff.} & & 2 & & 3 & & 4 & & 5 & & 6 & & \hdots\end{array}$