Thanks for the file.--- In

mathforfun@yahoogroups.com, adh_math

<no_reply@y...> wrote:

> --- 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,

> adh