A proof that the sum of two consecutive triangular numbers is square.

Without further ado we begin.

Take a look at the sequence 0, 1, 3, 6, 10, 15 ,21,…

Sum every two consecutive terms in the sequence.

What do you notice?

The sequence is the square numbers I hear you say.

And indeed they seem to be the proverbial square numbers i.e 1, 4, 9 ,16, 25, 36,… (Appears everywhere don’t you think?)

Proving this is very elementary starting with the fact that the nth triangular number is of the form x(x+1)/2 and that the (n-1)th triangular number is simply x(x-1)/2.

First,we expand both identities establishing that x(x+1)/2 is equal to(x^2+x)/2 (by the distributive property) and that x(x-1)/2 is equal to (x^2-x)/2 (by the distributive property again).

Next, we sum the the two like terms cancelling both the x and -x leaving(2x^2)/2 at the end.

Finally, we divide 2x^2 by 2 revealing x^2 at the end.

QED

Thank you,

Nunghead

PS: Did you catch that I alluded to the fact that the sum of n odd numbers is a perfect square. This is the reasoning for saying “Appears everywhere, don’t you think”.