How do you find three consecutive binomial coefficients in the relationship #1:2:3#?

Question

They are #((14),(4))#, #((14),(5))#, and #((14),(6))#, but I'd like to know how to obtain that result without using Pascal's Triangle.

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Answer 1 · 2016-11-22T00:55:46

We can represent the desired binomial coefficients as

#((n),(k)) = (n!)/(k!(n-k)!)#
#((n),(k+1))=(n!)/((k+1)!(n-k-1)!)#
#((n),(k+2))=(n!)/((k+2)!(n-k-2)!)#

Then, based on the given ratios, we have

#(((n),(k+1)))/(((n),(k)))= (k!(n-k)!)/((k+1)!(n-k-1)!) = (n-k)/(k+1) = 2#

#=>n-k = 2k+2#

#=> n-3k = 2#

and

#(((n),(k+2)))/(((n),(k+1)))=((k+1)!(n-k-1)!)/((k+2)!(n-k-2)!) = (n-k-1)/(k+2) = 3/2#

#=> 2n-2k-2=3k+6#

#=> 2n-5k=8#

#{(n-3k = 2),(2n-5k=8):}#

Solving this, we arrive at

#{(n = 14), (k=4):}#

Thus, our binomial coefficients are

#((14),(4)), ((14),(5)), ((14),(6))#

as expected.