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Math Help - Triangular \ Square number

  1. #1
    yakirtal76
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    Triangular \ Square number

    Hi all - can someone please help with the following

    Let Tm, Tn be triangular numbers.

    I need to show that (2m+1)^2*Tn + Tm is also triangular

    Now,,,,I have tried time and time again to expand the number and then somehow factorise it to make another triangular number , but with no success.

    (1) (2m+1)^2 = 4m^2 + 4m + 1
    (2) Tn = n(n+1) / 2
    (3) Tm = m(m+1) / 2

    But trying to multiply (1)*(2) and then add the term for (3) just ends up with a very long result, which I can't factorise.

    Any ideas please ??? Thanks a lot if anyone can explain it to me

    ps : (2m+1)^2 is of course a square number, S(2m+1), but am not aware of any property for the product of a square number with a triangular number....unless I am missing something??
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by yakirtal76
    Hi all - can someone please help with the following

    Let Tm, Tn be triangular numbers.

    I need to show that (2m+1)^2*Tn + Tm is also triangular

    Now,,,,I have tried time and time again to expand the number and then somehow factorise it to make another triangular number , but with no success.

    (1) (2m+1)^2 = 4m^2 + 4m + 1
    (2) Tn = n(n+1) / 2
    (3) Tm = m(m+1) / 2

    But trying to multiply (1)*(2) and then add the term for (3) just ends up with a very long result, which I can't factorise.

    Any ideas please ??? Thanks a lot if anyone can explain it to me

    ps : (2m+1)^2 is of course a square number, S(2m+1), but am not aware of any property for the product of a square number with a triangular number....unless I am missing something??
    I haven't worked it out all the way, but I'm getting a bit of success using an induction type proof, starting with the m's. I didn't finish it by working on the n's, but it shouldn't be too hard to complete. Maybe not, but it's a thought.

    -Dan
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  3. #3
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    We have (2m+1)^2 T_n + T_m = (2m+1)^2.n(n+1)/2 + m(m+1)/2. If this is a triangular number T_x then x(x+1) = (2m+1)^2.n(n+1) + m(m+1). The equation x(x+1) = F is quadratic in x with solution \frac{-1 \pm \sqrt{1+4F}}{2}. In this case F = (2m+1)^2.n(n+1) + m(m+1) and 4F+1 = (2m+1)^2(2n+1)^2. We find that x = \frac{(2m+1)(2n+1)-1}{2}.
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