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,r-1)
y=C(n,r) = n!/(r!(n-r)!)
z=C(n,r+1)

Setting 2x=y gives
n=3r-1
(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=14
and 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 2015-01-20 09:13:44