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I am a high school student, and while learning about figurate numbers, I came up with a relationship between pentagonal numbers, square numbers, and triangular numbers. I’m wondering if this formula is already known or if I discovered something new. Here’s what I found:

I discovered that the $n$-th pentagonal number $P_n$ can be expressed as the sum of the $n$-th square number $S_n$ and the $(n-1)$-th triangular number $T_{n-1}$: $$P_n = S_n + T_{n-1}$$

\begin{align} P_n &= \frac{n(3n - 1)}{2} &&\text{(pentagonal number)} \\[4pt] S_n &= n^2 &&\text{(square number)} \\[4pt] T_{n-1} &= \frac{(n - 1)n}{2} &&\text{(triangular number)} \end{align}

I’m wondering if this formula has already been discovered or if I am the first one to find it. I’d really appreciate any input or references!

Blue
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김준형
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1 Answers1

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I just want to add a proof without words:

enter image description here

(For people who don't see it, above the blue line is the triangle, and below is the square.)

ultralegend5385
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    This proof even generalizes for higher polygonal numbers! In general, we have $$P^s_n = P^{s-1}n + P^3{n-1}$$ where $P^s_n$ denotes the $n$th $s$-gonal number. – Alma Arjuna Mar 19 '25 at 18:27