Show that no three consecutive binomial coefficients can be in GP .

Explanation

Suppose that rth, (r + 1) th and (r + 2) th coefficients of
` (1 + x)^(n)` are in HP ,
i.e., ` ""^(n)C_(r-1),""^(n)C_(r),""^(n)C_(r+1)` are in HP .
Then , `(2)/(""^(n)C_(r)) = (1)/(""^(n)C_(r-1)) + (1)/(""^(n)C_(r+1))`
`rArr 2= (""^(n)C_(r))/(""^(n)C_(r-1))=(""^(n)C_(r))/(""^(n)C_(r+1)) [because (""^(n)C_(r))/(""^(n)C_(r-1))= (n-r+1)/(r)]`
` rArr 2= (n-r+1)/(r) + (r +1)/(n-r)`
`rArr 2r (n-r) = (n-r+1)(n-r)+ r(r+1)`
` rArr 2nr - 2r^(2) = n^(2) - nr - nr + r^(2) + n- r + r^(2) + r `
` rArr n^(2) - 4nr + 4r^(2) + n= 0 rArr (n-2r)^(2) + n= 0 `
which is not possible , as `(n-2r)^(2) ge 0 ` and n is a
positive integer .