## Re: Relation between nth and (2n)th Square Tri-numbers

Expand Messages
• Thanks for the file.--- In mathforfun@yahoogroups.com, adh_math ... file
Message 1 of 3 , Sep 11, 2005
Thanks for the file.--- In mathforfun@yahoogroups.com, adh_math
> --- In mathforfun@yahoogroups.com, "ramsey2879" <ramseykk2@a...>
> wrote:
> > A known recursive formula for square triangular numbers
> > (of the form T_k= k/2*{k+1}) is K_n = 34K_{n-1}-K_{n-2}+2.
> >
> > The following non-recursive formula also gives nth Square
> > Triangular number in terms of variable n.
> >
> > K_n = ({1 +sqrt[2]}^2n  {1-sqrt[2]}^2n)^2 / (4sqrt[2])^2
> >
> > Now from inspection it appears that given the square
> > triangular number K_n, the square triangular number
> > K_2n = T_(8*K_n), but I have difficulty using the non-recursive
> > formula for K_n and K_2n to verify this. I just thought that
> > some of you might like to try verifying this relationship for
> > yourself. Cheers
> >
>
> Perhaps you've found an answer in the meantime, but the uploaded
file
> tri_sq_identity.pdf contains a proof along the lines you seek.
>
> Regards,