Difference Between Adjacent Polygonal Numbers is Triangular Number
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Theorem
Let $\map P {k, n}$ be the $n$th $k$-gonal number.
Then:
- $\map P {k + 1, n} - \map P {k, n} = T_{n - 1}$
where $T_n$ is the $n$th triangular number.
Proof
From Closed Form for Polygonal Numbers:
- $\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Thus:
\(\ds \map P {k + 1, n} - \map P {k, n}\) | \(=\) | \(\ds \dfrac n 2 \paren {\paren {\paren {k + 1} - 2} n - \paren {k + 1} + 4} - \dfrac n 2 \paren {\paren {k - 2} n - k + 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac n 2 \paren {\paren {k - 1} n - k + 3} - \dfrac n 2 \paren {\paren {k - 2} n - k + 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac n 2 \paren {k n - n - k + 3} - \dfrac n 2 \paren {k n - 2 n - k + 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac n 2 \paren {k n - n - k + 3 - k n + 2 n + k - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac n 2 \paren {- n + 3 + 2 n - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n \paren {n - 1} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds T_{n - 1}\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $45$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $45$