# Triangular \ Square number

• Mar 26th 2006, 10:21 AM
yakirtal76
Triangular \ Square number

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??
• Mar 26th 2006, 04:12 PM
topsquark
Quote:

Originally Posted by yakirtal76

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
• Mar 30th 2006, 09:10 PM
rgep
We have $\displaystyle (2m+1)^2 T_n + T_m = (2m+1)^2.n(n+1)/2 + m(m+1)/2$. If this is a triangular number $\displaystyle T_x$ then $\displaystyle x(x+1) = (2m+1)^2.n(n+1) + m(m+1)$. The equation $\displaystyle x(x+1) = F$ is quadratic in $\displaystyle x$ with solution $\displaystyle \frac{-1 \pm \sqrt{1+4F}}{2}$. In this case $\displaystyle F = (2m+1)^2.n(n+1) + m(m+1)$ and $\displaystyle 4F+1 = (2m+1)^2(2n+1)^2$. We find that $\displaystyle x = \frac{(2m+1)(2n+1)-1}{2}$.