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Triangular numbers, proving numbers are tringular

  1. May 13, 2014 #1
    I've been learning about polygonal numbers, and one of the exercises in this book ask me to show that 9t[itex]_{n}[/itex]+1 [Fermat], 25t[itex]_{n}[/itex]+3, and 49t[itex]_{n}[/itex]+6 [both from Euler] are triangular numbers. I don't know how to approach these proofs, I've tried to show that they have some form similar to [itex]\frac{n(n+1)}{2}[/itex], but with no avail. But it looks like there is a pattern, that would be
    (2n+1)[itex]^{2}[/itex]t[itex]_{\alpha}[/itex]+[itex]\frac{n(n+1)}{2}[/itex], but I have no way of proving this. Could someone point me in the correct direction?

    t[itex]_{n}[/itex] and t[itex]_{\alpha}[/itex] are both triangular numbers.
    Last edited: May 13, 2014
  2. jcsd
  3. May 13, 2014 #2
    ##9\cdot\frac{n(n+1)}{2}+1=\frac{(3n+1)(3n+2)}{2}=\frac{(3n+1)\big((3n+1)+1 \big)}{2}##, yes?
  4. May 13, 2014 #3
    Wow, I was over thinking this a lot. Thank you, I can check my pattern with this too.
  5. May 13, 2014 #4
    Another one is ##5929t_n+741## is triangular whenever ##t_n## is triangular [gopher_p].

    But it looks like you've already figured out that pattern.
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