Problem:
In Pascal's triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
Row 0:Row 1:Row 2:Row 3:Row 4:Row 5:Row 6:111615141513101262013101415151611
In which row of Pascal's triangle do three consecutive entries occur that are in the ratio 3:4:5?
Solution:
Row n of Pascal's triangle consists of the binomial coefficients (kn),k=0,1,…,n. If three consecutive entries in row n of Pascal's triangle are in the ratio 3:4:5, then there is a positive integer k for which
43=(kn)(k−1n)=k!(n−k)!n!(k−1)!(n−k+1)!n!=(k−1)!(n−k+1)!k!(n−k)!=n−k+1k
and
54=(k+1n)(kn)=n!k!(n−k)!n!=(k+1)!(n−k−1)!(k+1)!(n−k−1)!=n−kk+1
It follows that
3n−7k=−3 and 4n−9k=5
Solving simultaneously gives k=27 and n=62. Thus, the consecutive entries (2662), (2762),(2862) in row 62 of Pascal's triangle are in the ratio 3:4:5.
The problems on this page are the property of the MAA's American Mathematics Competitions