Rows in Pascal's Triangle containing Numbers in Geometric Sequence

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Theorem

There exist no rows of Pascal's triangle which contain $3$ integers in geometric sequence.


Proof

Suppose $\dbinom n k$, $\dbinom n {k + 1}$ and $\dbinom n {k + 2}$ are in a geometric sequence.

Then:

\(\ds \dbinom n {k + 2} / \dbinom n {k + 1}\) \(=\) \(\ds \dbinom n {k + 1} / \dbinom n k\) Definition of Geometric Sequence
\(\ds \paren {\frac {n!} {\paren {n - k - 2}! \paren {k + 2}!} } \paren {\frac {\paren {n - k - 1}! \paren {k + 1}!} {n!} }\) \(=\) \(\ds \paren {\frac {n!} {\paren {n - k - 1}! \paren {k + 1}!} } \paren {\frac {\paren {n - k}! \paren k!} {n!} }\) Definition of Binomial Coefficient
\(\ds \frac {n - k - 1} {k + 2}\) \(=\) \(\ds \frac {n - k} {k + 1}\)
\(\ds \paren {n - k - 1} \paren {k + 1}\) \(=\) \(\ds \paren {n - k} \paren {k + 2}\)
\(\ds n k - k^2 - k + n -k - 1\) \(=\) \(\ds n k + 2 n - k^2 - 2 k\)
\(\ds n\) \(=\) \(\ds -1\)

Since $n \ge 0$, no row of Pascal's triangle contains $3$ integers in geometric sequence.


However, suppose one extends the definition of binomial coefficients to allow $n < 0$.

Then by Negated Upper Index of Binomial Coefficient, we have:

$\dbinom {-1} k = \paren {-1}^k$

which indeed forms a geometric sequence.

$\blacksquare$


Also see


Historical Note

This result, along with Rows in Pascal's Triangle containing Numbers in Arithmetic Sequence and Rows in Pascal's Triangle containing Numbers in Harmonic Sequence, was apparently published by Theodore Samuel Motzkin in Volume $12$ of Scripta Mathematica, but details have not been established.


Sources

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