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:​1​1​16​15​1415​1310​12620​1310​1415​15​16​1​1​
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