19. (CAN 5) The triangular array $\left(a_{n, k}\right)$ of numbers is given by $a_{n, 1}=1 / n$, for $n=1,2, \ldots, a_{n, k+1}=a_{n-1, k}-a_{n, k}$, for $1 \leq k \leq n-1$. Find the harmonic mean of the 1985th row.
19. First, we shall prove that the numbers in the $n$th row are exactly the numbers 
$$ \frac{1}{n\binom{n-1}{0}}, \frac{1}{n\binom{n-1}{1}}, \frac{1}{n\binom{n-1}{2}}, \ldots, \frac{1}{n\binom{n-1}{n-1}} $$
The proof of this fact can be done by induction. For small $n$, the statement can be easily verified. Assuming that the statement is true for some $n$, we have that the $k$th element in the $(n+1)$st row is, as is directly verified,
$$ \frac{1}{n\binom{n-1}{k-1}} - \frac{1}{(n+1)\binom{n}{k-1}} = \frac{1}{(n+1)\binom{n}{k}} $$
Thus (1) is proved. Now the geometric mean of the elements of the $n$th row becomes:
$$ \frac{1}{n \sqrt[n]{\binom{n-1}{0} \cdot \binom{n-1}{1} \cdots \binom{n-1}{n-1}}} \geq \frac{1}{n\left(\frac{\binom{n-1}{0} + \binom{n-1}{1} + \cdots + \binom{n-1}{n-1}}{n}\right)} = \frac{1}{2^{n-1}} $$
The desired result follows directly from substituting $n=1984$.