Loading ...
Sorry, an error occurred while loading the content.
 

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

Expand Messages
  • ramsey2879
    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
      <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
    Your message has been successfully submitted and would be delivered to recipients shortly.