Incidentally, the formula for the nth triangular number is actually 1/2*n*(n+1) rather than the 5*n*(n+1) given above.

Now we *can* prove that the sum of two consecutive triangular numbers is a square number:

1/2*n*(n+1) + 1/2*(n+1)*(n+2)
=1/2*(n+1)*(2n+2)
=(n+1)^2

Thought I was going nuts for a second there :)

In reply to Re: Triangle Numbers by Jobby
in thread Triangle Numbers by YuckFoo

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