Triangular numbers, proving numbers are tringular

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The discussion centers on proving that specific expressions, such as 9t_{n}+1, 25t_{n}+3, and 49t_{n}+6, represent triangular numbers. The initial challenge was to relate these expressions to the formula for triangular numbers, \(\frac{n(n+1)}{2}\). A breakthrough occurred when a participant identified a pattern involving \((2n+1)^{2}t_{\alpha}+\frac{n(n+1)}{2}\). This led to the realization that 9 times the triangular number formula plus one can be expressed in a triangular number format. The conversation concludes with an acknowledgment that the pattern has been successfully identified and verified.
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I've been learning about polygonal numbers, and one of the exercises in this book ask me to show that 9t_{n}+1 [Fermat], 25t_{n}+3, and 49t_{n}+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 \frac{n(n+1)}{2}, but with no avail. But it looks like there is a pattern, that would be
(2n+1)^{2}t_{\alpha}+\frac{n(n+1)}{2}, but I have no way of proving this. Could someone point me in the correct direction?

t_{n} and t_{\alpha} are both triangular numbers.
 
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##9\cdot\frac{n(n+1)}{2}+1=\frac{(3n+1)(3n+2)}{2}=\frac{(3n+1)\big((3n+1)+1 \big)}{2}##, yes?
 
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Wow, I was over thinking this a lot. Thank you, I can check my pattern with this too.
 
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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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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