October
30, 2006
Does
any row of Pascal's triangle contain three consecutive entries that are
in the ratio 1:2:3? Recall that the $n$th row ($n=0,1,2,\dots$)
consists of the entries
$\displaystyle\left ( \begin{array}{c} n \\
k \end{array}\right )=\frac{n!}{k!(n-k)!} \quad (k=0,1,2,\dots, n)$
Solution. Three
consecutive entries of the $n$th row, say
$\displaystyle\left ( \begin{array}{c} n \\
k \end{array}\right )$, $\displaystyle\left ( \begin{array}{c} n \\
k+1 \end{array}\right )$, and $\displaystyle\left ( \begin{array}{c} n
\\
k+2 \end{array}\right )$,
stand in the ratio 1:2:3 if and only if
$\displaystyle\frac{n-k}{k+1}=2$ and
$\displaystyle\frac{n-k-1}{k+2}=\frac32$.
These equations, obtained by expanding the binomial coefficients as in
the problem statement, can be rewritten as $n=3k+2$ and $2n=5k+8$.
There is exactly one solution to these two equations, namely, $n=14$
and $k=4$.