laziness, impatience, and hubris PerlMonks

### Re: Triangle Numbers

by Jobby (Monk)
 on Jul 26, 2002 at 01:36 UTC ( #185395=note: print w/replies, xml ) Need Help??

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 :)

Replies are listed 'Best First'.
Re: Re: Triangle Numbers
by Anonymous Monk on Jul 26, 2002 at 01:40 UTC
You missed the decimal point in front of the 5.
which one is right

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