Element of Leibniz Harmonic Triangle is Sum of Numbers Below

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Theorem

The elements in the Leibniz harmonic triangle are the sum of the elements immediately below them.


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Proof

By definition of Leibniz harmonic triangle, element $\tuple {n, m}$ is:

$\dfrac 1 {\paren {n + 1} \binom n m}$


Thus we have:

\(\ds \) \(\) \(\ds \dfrac 1 {\paren {n + 2} \binom {n + 1} m} + \dfrac 1 {\paren {n + 2} \binom {n + 1} {m + 1} }\) Elements of Leibniz Harmonic Triangle immediately below
\(\ds \) \(=\) \(\ds \dfrac {\binom {n + 1} {m + 1} + \binom {n + 1} m} {\paren {n + 2} \binom {n + 1} m \binom {n + 1} {m + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {\binom {n + 2} {m + 1} } {\paren {n + 2} \binom {n + 1} m \binom {n + 1} {m + 1} }\) Pascal's Rule
\(\ds \) \(=\) \(\ds \dfrac {\frac {\paren {n + 2}!} {\paren {m + 1}! \paren {n - m + 1}!} } {\paren {n + 2} \frac {\paren {n + 1}!} {m! \paren {n - m + 1}!} \frac {\paren {n + 1}!} {\paren {m + 1}! \paren {n - m}!} }\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds \dfrac {\paren {n + 1}!} {\frac {\paren {n + 1}!} {m!} \frac {\paren {n + 1}!} {\paren {n - m}!} }\) some simplification
\(\ds \) \(=\) \(\ds \dfrac 1 {\frac {\paren {n + 1}!} {m! \paren {n - m}!} }\) further simplification
\(\ds \) \(=\) \(\ds \dfrac 1 {\paren {n + 1} \frac {n!} {m! \paren {n - m}!} }\) further simplification
\(\ds \) \(=\) \(\ds \dfrac 1 {\paren {n + 1} \binom n m}\) Definition of Binomial Coefficient

Hence the result.

$\blacksquare$


Sources

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