If the 3rd, 4th , 5th and 6th term in the expansion of `(x+alpha)^n` be, respectively, `a ,b ,ca n dd ,` prove that `(b^2-a c)/(c^2-b d)=(5a)/(3c)dot`

Explanation

We know that
`(T_(r+1))/(T_(r)), (n-r+1)/(r)(alpha)/(x)`
`T_(3)=a,T_(4)=b,T_(5)=c,T_(6)=d`
Putting `r = 3, 4,5` in the abvoe we get
`(T_(4))/(T_(3)) = (n-2)/(3)(alpha)/(x) = b/a`
`(T_(5))/(T_(4)) = (n-3)/(4)(alpha)/(x) = c/b`
`(T_(6))/(T_(5)) = (n-4)/(5) (alpha)/(x) = d/c`
We have to prove
`(b^(2)-ac)/(c^(2)-bd) = (5a)/(3c)`
or `((b)/(c)-(a)/(b))/((c)/(b)-(d)/(c)) = (5a)/(3c)`
Now, `((b)/(c) - (a)/(b))/((c)/(b)-(d)/(c)) = ((4x)/((n-3)alpha)-(3x)/((n-2)alpha))/(((n-3)alpha)/(4x) - ((n-4)alpha)/(5x))`
`= (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))(4n-8-3n+9)/(5n-15-4n+16)`
`= (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))`
Also, `(5a)/(3c) = (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))`
`:. L.H.S. = R.H.S.`