Does any row of a Pascal's triangle have three consecutive entries that are in the ratio 1:2:3 ?
The answer is yes. Entries 4, 5, 6 in row 14 are
1001, 2002, 3003.
Method: Write out the formulas for 3 consecutive numbers in a row:
x=C(n,r1)
y=C(n,r) = n!/(r!(nr)!)
z=C(n,r+1)
Setting 2x=y gives
n=3r1(this gives an infinity of solutions such as r=2 the 5 and 10 in row 5)
Setting 3y=2z gives
n=2.5r+1.5(this gives an infinity of solutions such as r=1 the 4 and 6 in row 5)
Create a system with the two bold formulas above gives the solution
r=5 n=14and indeed if we check the numbers
C(14,4)=1001
C(14,5)=2002
C(14,6)=3003
The solution found implies there are no others.

Posted by Jer
on 20150120 09:13:44 