jmc

Question

In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio

3 : 4 : 5

?

(The top row of Pascal's Triangle has only a single

1

and is the

0

th row.)

Accepted Answer

Let the three entries be nr, nr+1, and nr+2, respectively. Then we have nrnr+1 = 34 and nr+1nr+2 = 45.We simplify the left-hand side of the first equation: nrnr+1 = n!r!(n-r)!n!(r+1)!)(n-r-1)! = n!r!(n-r)! (r+1)!(n-r-1)!n! = r+1n-r.Therefore, r+1n-r = 34. Similarly, the second equation becomes r+2n-r-1 = 45. Cross-multiplying in both equations, we have 4r+4 = 3n-3r and 5r+10 = 4n-4r-4.Solving for r in the first equation gives r = 3n-47, and then we have 9(3n-47) + 14 = 4n,and solving for n gives n = 62.