# Rows in Pascal's Triangle containing Numbers in Harmonic Sequence

## Theorem

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

## Proof

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

Then:

 $\ds \dbinom n {k + 2}^{-1} - \dbinom n {k + 1}^{-1}$ $=$ $\ds \dbinom n {k + 1}^{-1} - \dbinom n k^{-1}$ Definition of Harmonic Sequence $\ds \frac {\paren {n - k - 2}! \paren {k + 2}!} {n!} - \frac {\paren {n - k - 1}! \paren {k + 1}!} {n!}$ $=$ $\ds \frac {\paren {n - k - 1}! \paren {k + 1}!} {n!} - \frac {\paren {n - k}! \paren k!} {n!}$ Definition of Binomial Coefficient $\ds \paren {k + 2} \paren {k + 1} - \paren {n - k - 1} \paren {k + 1}$ $=$ $\ds \paren {n - k - 1} \paren {k + 1} - \paren {n - k} \paren {n - k - 1}$ Multiply both sides by $\dfrac {n!} {\paren {n - k - 2}! \paren k!}$ $\ds k^2 + 3 k + 2 - n k + k^2 + k - n + k + 1$ $=$ $\ds n k - k^2 - k + n - k - 1 - n^2 + n k + n + k n - k^2 - k$ $\ds n^2 - \paren {4 k + 3} n + \paren {4 k^2 + 8 k + 4}$ $=$ $\ds 0$

This is a quadratic equation in $n$, so we can calculate its discriminant.

Notice that for each $k \ge 0$:

$\paren {4 k + 3}^2 - 4 \paren {4 k^2 + 8 k + 4} = - \paren {8 k + 7} < 0$

By Solution to Quadratic Equation with Real Coefficients, there is no real solution for $n$.

Therefore there is no row of Pascal's triangle which contain $3$ integers in harmonic sequence.

$\blacksquare$

## 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 Geometric Sequence, was apparently published by Theodore Samuel Motzkin in Volume $12$ of Scripta Mathematica, but details have not been established.