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

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.