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

let `t_3 = a`

`t_4 = b`

`t_5 = c`

`t_6 = d`

for the expansion `(x+ alpha)^n`

`t_(r+1)/t_r= (n-r+1)/r* alpha/x`

`t_4/t_3= (n-3+1)/3* alpha/x = b/a = (n-2)/3* alpha/x`

`t_5/t_4= (n-4+1)/4* alpha/x = c/b = (n-3)/4* alpha/x `

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